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Solve quadratic equations: word problems

Date\newlineName\newlineFind the perimeter and area of the shaded section. Fill in each □ \square .\newline(11)\newlinePerimeter of the shaded section\newline= = Circumference of the 16 cm 16 \mathrm{~cm} circle + Circumference of the 8 cm 8 \mathrm{~cm} circle\newline=(□×3.14)+(□×3.14) =(\square \times 3.14)+(\square \times 3.14) \newline= = \newline□ \square + + □ \square = = □ \square cm \mathrm{cm} \newlineArea of the shaded section\newline= = Area of the 16 cm 16 \mathrm{~cm} circle - Area of the 8 cm 8 \mathrm{~cm} circle\newline=(□×□×3.14)−(□×□×3.14)=□−□cm2 \begin{array}{l} =(\square \times \square \times 3.14)-(\square \times \square \times 3.14) \\ =\square-\square \mathrm{cm}^{2} \end{array} \newline33
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Here are the answers: The net present value (NPV) of the relevant cash flows associated with the lease option is: Initial investment: \$(\(-350\),\(000\)) (transforming the leased space into an ice cream shop) Annual rent: \$(\(-36\),\(000\)) \times \(3\).\(433\) (discount factor for \(5\) years at \(14\)\% discount rate) = \$(\(-123\),\(948\)) Total NPV: \$(\(-350\),\(000\)) - \$(\(-123\),\(948\)) = \$(\(-473\),\(948\)) The NPV of the relevant cash flows associated with the buy-and-build option is: Initial investment: \$(\(-1\),\(200\),\(000\)) (land, building, and paved parking lot) Annual excess operating costs: \$(\(-15\),\(000\)) \times \(3\).\(433\) (discount factor for \(5\) years at \(14\)\% discount rate) = \$(\(-51\),\(495\)) Terminal value: \$\(1\),\(300\),\(000\) / (\(1\) + \(0\).\(14\))^\(5\) = \$\(833\),\(581\) (present value of the market value at the end of \(5\) years) Total NPV: \$(\(-1\),\(200\),\(000\)) - \$(\(-51\),\(495\)) + \$\(833\),\(581\) = \$(\(-417\),\(914\)) To make Annie's indifferent between the lease and buy-and-build alternatives, the market value of the commercial property at the end of \(5\) years would need to be: \$\(1\),\(200\),\(000\) (initial investment) + \$\(51\),\(495\) (excess operating costs) - \$\(350\),\(000\) (initial investment for lease option) = \$\(901\),\(495\) Then, discount this amount to its present value using the \(14\)\% discount rate and \(5\)-year time horizon: \$\(901\),\(495\) / (\(1\) + \(0\).\(14\))^\(5\) = \$\(574\),\(939\) So, the market value of the commercial property at the end of \(5\) years would need to be at least \$\(574\),\(939\) for Annie's to be indifferent between the lease and buy-and-build options.
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Three identical units of merchandise were purchased during March, as follows:\newline\begin{tabular}{rlcr} \newline& Steele Plate & Units & Cost \\\newline\hline Mar. 33 & Purchase & 11 & $830 \$ 830 \\\newline1010 & Purchase & 11 & 840840 \\\newline1919 & Purchase & 11 & 880880 \\\newline\cline { 22 - 33 } & Total & 33 & $2,550 \$ 2,550 \\\newline\hline\newline\end{tabular}\newlineAssume that one unit is sold on March 2323 for $1,125 \$ 1,125 . Determine the gross profit for March and ending inventory on March 3131 using the (a) FIFO, (b) LIFO, and (c) weighted average cost methods.\newline\begin{tabular}{lll} \newline& Gross Profit & Ending Inventory \\\newline\hline a. First-in, first-out (FIFO) & $□ \$ \square \\\newlineb. Last-in, first-out (LIFO) & $□ \$ \square \\\newlinec. Weighted average cost & $□ \$ \square \newline\end{tabular}
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88. The community garden club has a vegetable garden that measures 15 m 15 \mathrm{~m} by 30 m 30 \mathrm{~m} . One of the members has donated a new piece of land for a larger garden. They plan to increase the garden by 250 m2 250 \mathrm{~m}^{2} . However, because of the dimensions of the new land, both dimensions of the original garden must be increased by the same amount. Determine the dimensions of the new garden.\newline [20 m 35 m ]  \text { [20 m } 35 \mathrm{~m} \text { ] }
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Soal 11 (Nilai =25 =25 )\newlineBerikut adalah data kuantitas produksi dan harga per unit produk yang dihasilkan oleh PT. IMB\newline\begin{tabular}{|l|l|l|l|l|l|l|}\newline\hline \multirow{22}{*}{ Produk } & \multicolumn{22}{|c|}{ KuantitasProduksi (ribuan unit) } & \multicolumn{33}{c|}{ Harga per satuan (rupiah) } \\\newline\cline { 22 - 77 } & 20162016 & 20172017 & 20182018 & 20162016 & 20172017 & 20182018 \\\newline\hline Balpoint & 100100 & 150150 & 250250 & 22.500500 & 22.700700 & 33.000000 \\\newline\hline Pensil & 150150 & 200200 & 250250 & 11.500500 & 11.700700 & 11.900900 \\\newline\hline\newline\end{tabular}\newline\begin{tabular}{|l|l|l|l|l|l|l|}\newline\hline Penggaris & 125125 & 150150 & 175175 & 22.000000 & 22.200200 & 22.400400 \\\newline\hline Tipe-X & 130130 & 135135 & 140140 & 55.000000 & 55.200200 & 55.400400 \\\newline\hline Penghapus & 135135 & 140140 & 150150 & 11.000000 & 11.200200 & 11.500500 \\\newline\hline\newline\end{tabular}\newlineSaudara diminta:\newline11. Menghitung indeks harga relative sederhana semua jenis produk tahun 20172017 dengan tahun dasar 20162016\newline22. Menghitung indeks harga agregat sederhana tahun 20182018 dengan tahun dasar 20162016\newline33. Menghitung indeks rata-rata relatif tahun 20182018 dengan tahun dasar 20172017\newline44. Menghitung indeks agregat tertimbang tahun 20182018 dengan tahun dasar 20162016 dengan rumus Fisher, Drobisch, Marshal-Edgeworth, dan Wals
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Dr. Rollins is both an anthropologist and archeologist. While excavating some ruins in South America, he discovered a scale drawing of a replica of a Mayan pyramid.\newline- The scale for the drawing to the replica was 11 inch : 22 feet.\newline- The scale for the replica to the actual pyramid was 11 foot : 1414 feet.\newlinethe height of the pyramid on the drawing was 312 3 \frac{1}{2} inches, what was the height of the actual pyramid?\newlineA 196196 feet\newlineB 9898 feet\newlineC 4949 feet\newlineD 9191 feet\newlineLearning 20242024, All Rights Reserved. Terms | Privacy PHONE 11−877-877.377377−9537-9537 | FAX 11−877-877−816-816−0808-0808 Blog
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Name:\newline \qquad \newlineLESSON 1010 SESSION 22\newlinePractice Adding Decimals\newlineStudy the Example showing decimal addition using a place-value chart. Then solve problems 11−5-5.\newlineEXAMPLE\newlineAlana walks 33.4545 miles before lunch and 55.1818 miles after lunch. How many miles does she walk in all?\newline\begin{tabular}{|c|c|c|c|}\newline\hline Ones &. & Tenths & Hundredths \\\newline\hline 33 &. & 44 & 55 \\\newline\hline 55 &. & 11 & 88 \\\newline\hline\newline\end{tabular}\newline33 ones +55 ones =8 =8 ones\newline44 tenths +11 tenth =5 =5 tenths\newline55 hundredths +88 hundredths =13 =13 hundredths\newline Sum =8 ones +5 tenths +13 hundredths =8 ones +6 tenths +3 hundredths  \begin{aligned} \text { Sum } & =8 \text { ones }+5 \text { tenths }+13 \text { hundredths } \\ & =8 \text { ones }+6 \text { tenths }+3 \text { hundredths } \end{aligned} \newline Sum =8 ones +5 tenths +13 hundredths  \text { Sum }=8 \text { ones }+5 \text { tenths }+13 \text { hundredths } \newline Sum =8 ones +5 tent  \text { Sum }=8 \text { ones }+5 \text { tent } \newline Sum =8 ones +5 tenth  \text { Sum }=8 \text { ones }+5 \text { tenth } \newlineAlana walks 88.6363 miles in all.\newline(11) Look at the Example. Suppose Alana walks 66.66 miles the next day. Complete the place-value chart and steps to find the number of miles she walks in two days.\newline\begin{tabular}{|c|c|c|c|}\newline\hline Ones &. & Tenths & Hundredths \\\newline\hline 88 &. & 66 & 33 \\\newline\hline &. & & \\\newline\hline\newline\end{tabular}\newlineSum =14 =14 ones + \qquad tenths + \qquad hundredths\newline=……... ones +2 tenths +3 hundredths  =\ldots \ldots . . . \text { ones }+2 \text { tenths }+3 \text { hundredths } \newlineones + tenths +\newlineAlana walks \qquad miles in two days.\newlineoCurriculum Associates, LLC\newlineCopying is not permitted.\newlineLesson 1010 Add Decimals\newline197197
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33) A={x∣ −3≤x<−12},B={x∣−1<x≤3} A=\left\{x \left\lvert\,-3 \leq x<-\frac{1}{2}\right.\right\}, B=\{x \mid-1<x \leq 3\} dan C={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} \newlinea. A∪B={x∣ −3≤x<−12∪−1<x≤3}=… A \cup B=\left\{x \left\lvert\,-3 \leq x<-\frac{1}{2} \cup-1<x \leq 3\right.\right\}=\ldots .\newlineb. A∩B={x∣ −3≤x<−12∩−1<x≤3}=… \quad A \cap B=\left\{x \left\lvert\,-3 \leq x<-\frac{1}{2} \cap-1<x \leq 3\right.\right\}=\ldots ...\newlinec. (A∪B)∩C=…={x∣ −3≤x≤3∩12<x<5}=… \quad(A \cup B) \cap C=\ldots=\left\{x \left\lvert\,-3 \leq x \leq 3 \cap \frac{1}{2}<x<5\right.\right\}=\ldots .\newlined. (A∩B)∪C=…={x∣ −1<x<−12∪12<x<5}=… (A \cap B) \cup C=\ldots=\left\{x \left\lvert\,-1<x<-\frac{1}{2} \cup \frac{1}{2}<x<5\right.\right\}=\ldots .\newline44) A={x∣x2−x−2=0},B={x∣x2−x−6=0} \quad A=\left\{x \mid x^{2}-x-2=0\right\}, B=\left\{x \mid x^{2}-x-6=0\right\} , dan C={x∣x2−4=0} C=\left\{x \mid x^{2}-4=0\right\} .\newlineA={x∣x2−x−2=(x+1)(x−2)=0⇒x1=−1;x2=2} A=\left\{x \mid x^{2}-x-2=(x+1)(x-2)=0 \Rightarrow x_{1}=-1 ; x_{2}=2\right\} , sehingga\newlineA={−1,2} A=\{-1,2\} .\newlineC={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 00, sehingga C={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 11.\newlineC={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 22, sehingga C={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 33.\newlineJadi,\newlinea. C={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 44.\newlineb. C={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 55.\newlinec. C={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 66.\newlined. C={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 77.\newlinee. C={x∣ 12<x<5} C=\left\{x \left\lvert\, \frac{1}{2}<x<5\right.\right\} 88
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The function h(t)=−16t2+144h(t) = -16t^{2} + 144 represents the height, h(t)h(t), in feet, of an object from the ground at tt seconds after it is dropped. A realistic domain (for time) for this function is:\newline(A)(A) −3≤t≤3-3 \leq t \leq 3\newline(B)(B) 0≤t≤30 \leq t \leq 3\newline(C)(C) 0≤h≤1440 \leq h \leq 144\newline(D)(D) all real numbers
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