Resources
Testimonials
Plans
Sign in
Sign up
Resources
Testimonials
Plans
Home
Math Problems
Grade 8
Solve equations with variables on both sides: fractional coefficients
Solve for
z
z
z
.
\newline
5
3
z
+
1
=
3
−
2
3
z
\frac{5}{3}z + 1 = 3 - \frac{2}{3}z
3
5
z
+
1
=
3
−
3
2
z
\newline
z
=
z =
z
=
____
Get tutor help
Solve for
f
f
f
.
\newline
2
−
1
4
f
=
f
−
3
4
f
+
3
2 - \frac{1}{4}f = f - \frac{3}{4}f + 3
2
−
4
1
f
=
f
−
4
3
f
+
3
\newline
f
=
f =
f
=
__
Get tutor help
classroom.google.com/c/
N
T
U
O
N
T
I
2
N
T
Y
NTUONTI2NTY
NT
U
ONT
I
2
NT
Y
\newline
Inc Student Logins - St. B Apple Music
\newline
tions: Find the value of
\newline
ϵ
\epsilon
ϵ
\newline
x
=
x=
x
=
\newline
y
=
y=
y
=
Get tutor help
Find five Rationd Numbers betwem
3
3
3
\newline
and
2
2
2
using
\newline
r
+
s
r+s
r
+
s
formule?
Get tutor help
Geo(#
3
3
3
Practice) - Unit
8
8
8
Review (Kite
P
Q
R
S
PQRS
PQRS
is a trapezoid. Find each measure.
m
/
_
Q
=
m/\_Q=
m
/_
Q
=
m
/
_
s
=
m/\_s=
m
/_
s
=
Get tutor help
Solve for the exact value of
x
x
x
.
\newline
2
ln
(
6
x
+
5
)
−
11
=
−
3
2 \ln (6 x+5)-11=-3
2
ln
(
6
x
+
5
)
−
11
=
−
3
\newline
Answer:
Get tutor help
Solve for the exact value of
x
x
x
.
\newline
6
ln
(
4
x
+
9
)
+
4
=
−
14
6 \ln (4 x+9)+4=-14
6
ln
(
4
x
+
9
)
+
4
=
−
14
\newline
Answer:
Get tutor help
Solve for
d
d
d
.
\newline
4
5
d
+
3
=
−
2
−
1
5
d
\frac{4}{5}d + 3 = -2 - \frac{1}{5}d
5
4
d
+
3
=
−
2
−
5
1
d
\newline
d
=
d =
d
=
__
Get tutor help
Solve for
s
s
s
.
\newline
5
8
−
3
8
s
=
2
−
s
\frac{5}{8} - \frac{3}{8}s = 2 - s
8
5
−
8
3
s
=
2
−
s
\newline
s
=
s =
s
=
__
Get tutor help
Y
=
4
7
x
Y=\frac{4}{7}x
Y
=
7
4
x
\newline
2
3
x
=
y
+
5
7
\frac{2}{3}x=y+\frac{5}{7}
3
2
x
=
y
+
7
5
Get tutor help
In
△
I
J
K
,
m
∠
I
=
(
x
+
6
)
∘
,
m
∠
J
=
(
2
x
+
16
)
∘
\triangle \mathrm{IJK}, \mathrm{m} \angle I=(x+6)^{\circ}, \mathrm{m} \angle J=(2 x+16)^{\circ}
△
IJK
,
m
∠
I
=
(
x
+
6
)
∘
,
m
∠
J
=
(
2
x
+
16
)
∘
, and
m
∠
K
=
(
6
x
+
14
)
∘
\mathrm{m} \angle K=(6 x+14)^{\circ}
m
∠
K
=
(
6
x
+
14
)
∘
. What is the value of
x
x
x
?
\newline
Answer:
Get tutor help
In
△
B
C
D
,
B
D
‾
\triangle \mathrm{BCD}, \overline{B D}
△
BCD
,
B
D
is extended through point
D
\mathrm{D}
D
to point
E
,
m
∠
C
D
E
=
(
6
x
+
7
)
∘
\mathrm{E}, \mathrm{m} \angle C D E=(6 x+7)^{\circ}
E
,
m
∠
C
D
E
=
(
6
x
+
7
)
∘
,
m
∠
B
C
D
=
(
3
x
+
9
)
∘
\mathrm{m} \angle B C D=(3 x+9)^{\circ}
m
∠
BC
D
=
(
3
x
+
9
)
∘
, and
m
∠
D
B
C
=
(
x
+
20
)
∘
\mathrm{m} \angle D B C=(x+20)^{\circ}
m
∠
D
BC
=
(
x
+
20
)
∘
. What is the value of
x
?
x ?
x
?
\newline
Answer:
Get tutor help
What is the amplitude of
\newline
g
(
x
)
=
−
9
cos
(
π
2
x
−
6
)
+
8
?
g(x)=-9 \cos \left(\frac{\pi}{2} x-6\right)+8 ?
g
(
x
)
=
−
9
cos
(
2
π
x
−
6
)
+
8
?
\newline
units
Get tutor help
What is the slope of the line?
\newline
2
x
+
4
y
=
6
x
−
y
2x+4y=6x-y
2
x
+
4
y
=
6
x
−
y
\newline
Choose
1
1
1
answer:
\newline
(A)
5
4
\frac{5}{4}
4
5
\newline
(B)
3
2
\frac{3}{2}
2
3
\newline
(C)
4
5
\frac{4}{5}
5
4
\newline
(D)
2
3
\frac{2}{3}
3
2
Get tutor help
