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Let’s check out your problem:
Rewrite in simplest terms:
−
0.5
(
−
3
k
+
8
)
−
0.4
(
−
5
k
+
3
)
-0.5(-3 k+8)-0.4(-5 k+3)
−
0.5
(
−
3
k
+
8
)
−
0.4
(
−
5
k
+
3
)
\newline
Answer:
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Math Problems
Algebra 1
Multiplication with rational exponents
Full solution
Q.
Rewrite in simplest terms:
−
0.5
(
−
3
k
+
8
)
−
0.4
(
−
5
k
+
3
)
-0.5(-3 k+8)-0.4(-5 k+3)
−
0.5
(
−
3
k
+
8
)
−
0.4
(
−
5
k
+
3
)
\newline
Answer:
Distribute coefficients:
Distribute the first negative coefficient
−
0.5
-0.5
−
0.5
to each term inside the first parentheses.
\newline
−
0.5
×
−
3
k
+
−
0.5
×
8
-0.5 \times -3k + -0.5 \times 8
−
0.5
×
−
3
k
+
−
0.5
×
8
Perform multiplication:
Perform the multiplication for each term.
\newline
1.5
k
−
4
1.5k - 4
1.5
k
−
4
Distribute coefficients:
Distribute the second negative coefficient
−
0.4
-0.4
−
0.4
to each term inside the second parentheses.
\newline
−
0.4
×
−
5
k
+
−
0.4
×
3
-0.4 \times -5k + -0.4 \times 3
−
0.4
×
−
5
k
+
−
0.4
×
3
Perform multiplication:
Perform the multiplication for each term.
2
k
−
1.2
2k - 1.2
2
k
−
1.2
Combine results:
Combine the results from Step
2
2
2
and Step
4
4
4
.
\newline
1.5
k
−
4
+
2
k
−
1.2
1.5k - 4 + 2k - 1.2
1.5
k
−
4
+
2
k
−
1.2
Combine like terms:
Combine like terms by adding the coefficients of
k
k
k
and the constant terms separately.
(
1.5
k
+
2
k
)
+
(
−
4
−
1.2
)
(1.5k + 2k) + (-4 - 1.2)
(
1.5
k
+
2
k
)
+
(
−
4
−
1.2
)
Perform addition:
Perform the addition for each group of like terms.
3.5
k
−
5.2
3.5k - 5.2
3.5
k
−
5.2
More problems from Multiplication with rational exponents
Question
Let
h
(
x
)
=
log
2
(
x
)
h(x)=\log _{2}(x)
h
(
x
)
=
lo
g
2
(
x
)
.
\newline
Can we use the mean value theorem to say the equation
h
′
(
x
)
=
1
h^{\prime}(x)=1
h
′
(
x
)
=
1
has a solution where
1
<
x
<
2
1<x<2
1
<
x
<
2
?
\newline
Choose
1
1
1
answer:
\newline
(A) No, since the function is not differentiable on that interval.
\newline
(B) No, since the average rate of change of
h
h
h
over the interval
1
≤
x
≤
2
1 \leq x \leq 2
1
≤
x
≤
2
isn't equal to
1
1
1
.
\newline
(C) Yes, both conditions for using the mean value theorem have been met.
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What is the sign of
39
11
13
+
(
−
41
1
13
)
?
39 \frac{11}{13}+\left(-41 \frac{1}{13}\right) \text { ? }
39
13
11
+
(
−
41
13
1
)
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Neither positive nor negative-the sum is zero.
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What is the sign of
−
1042
+
1042
?
-1042+1042 ?
−
1042
+
1042
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Neither positive nor negative-the sum is zero.
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Question
What is the sign of
−
18
9
17
+
(
−
18
9
17
)
-18 \frac{9}{17}+\left(-18 \frac{9}{17}\right)
−
18
17
9
+
(
−
18
17
9
)
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Neither positive nor negative-the sum is zero.
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Question
What is the sign of
23
45
+
(
−
23
45
)
?
\frac{23}{45}+\left(-\frac{23}{45}\right) ?
45
23
+
(
−
45
23
)
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Neither positive nor negative-the sum is zero.
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Question
What is the sign of
45
+
(
−
38
)
45+(-38)
45
+
(
−
38
)
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Neither positive nor negative-the sum is zero.
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Question
What is the sign of
−
49.8
+
61.2
-49.8+61.2
−
49.8
+
61.2
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Neither positive nor negative-the sum is zero.
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Question
What is the sign of
−
1.69
+
(
−
1.69
)
?
-1.69+(-1.69) ?
−
1.69
+
(
−
1.69
)?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Neither positive nor negative-the sum is zero.
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Question
What is the sign of
37
+
(
−
37
)
37+(-37)
37
+
(
−
37
)
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Neither positive nor negative-the sum is zero.
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Posted 9 months ago
Question
g
(
x
)
=
2
x
−
9
g
′
(
x
)
=
?
\begin{array}{l} g(x)=\sqrt{2 x-9} \\ g^{\prime}(x)=? \end{array}
g
(
x
)
=
2
x
−
9
g
′
(
x
)
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
2
x
−
9
2
\frac{\sqrt{2 x-9}}{2}
2
2
x
−
9
\newline
(B)
1
2
x
−
9
\frac{1}{\sqrt{2 x-9}}
2
x
−
9
1
\newline
(C)
1
2
2
x
−
9
\frac{1}{2 \sqrt{2 x-9}}
2
2
x
−
9
1
\newline
(D)
1
x
\frac{1}{\sqrt{x}}
x
1
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Posted 9 months ago
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