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Question1. If 2d=−1032d = \frac{-10}{3}, find dd. A. d=−2d=-2 B. d=2d=2 C. d=8d=8 D. d=10d=10
2. Expand (2k+3)(4k−5)(2k+3)(4k-5). A. 8k2+2k+158k^{2}+2k+15 B. 8k2+2k−158k^{2}+2k-15 C. 8k2−2k−158k^{2}-2k-15 D. 8k2+4k−158k^{2}+4k-15
3. Simplify 7n2−n(1−n)7n^{2}-n(1-n). A. 8n2−n8n^{2}-n B. 6n2−n6n^{2}-n C. 8n2+n8n^{2}+n D. 6n2+n6n^{2}+n
4. Solve t3−1<t\frac{t}{3}-1 < t. A. 13<t\frac{1}{3} < t B. −13<t-\frac{1}{3} < t C. −32<t\frac{-3}{2} < t D. 32<t\frac{3}{2} < t
5. Solve a+1>12a−2a+1 > \frac{1}{2}a-2. A. a>−6a > -6 B. a>6a > 6 C. a>23a > \frac{2}{3} D. a>−2a > -2

Studdy Solution

STEP 1

Assumptions1. For problem1, the given equation is d=−103d = \frac{-10}{3} and we need to solve for dd. . For problem, we need to expand and simplify the expression (k+3)(4k−5)(k+3)(4k-5).
3. For problem3, we need to simplify the expression 7n−n(1−n)7n^ - n(1-n).
4. For problem4, we need to solve the inequality t3−1<t\frac{t}{3}-1 < t.
5. For problem5, we need to solve the inequality a+1>1a−a+1 > \frac{1}{}a-.

STEP 2

For problem1, we need to isolate dd in the equation 2d=−102d = \frac{-10}{}. We can do this by dividing both sides of the equation by2.
d=2d2=−102d = \frac{2d}{2} = \frac{\frac{-10}{}}{2}

STEP 3

Now, calculate the value of dd.
d=−1032=−106=−53d = \frac{\frac{-10}{3}}{2} = \frac{-10}{6} = \frac{-5}{3}The correct answer for problem1 is not listed in the options.

STEP 4

For problem2, we need to expand the expression (2k+3)(4k−)(2k+3)(4k-). This can be done using the distributive property of multiplication over addition.
(2k+3)(4k−)=2k⋅4k+2k⋅−+3⋅4k+3⋅−(2k+3)(4k-) =2k \cdot4k +2k \cdot - +3 \cdot4k +3 \cdot -

STEP 5

Now, simplify the expression.
(2k+3)(4k−5)=8k2−10k+12k−15=8k2+2k−15(2k+3)(4k-5) =8k^2 -10k +12k -15 =8k^2 +2k -15The correct answer for problem2 is B. 8k2+2k−158k^2 +2k -15.

STEP 6

For problem3, we need to simplify the expression n2−n(1−n)n^2 - n(1-n). This can be done by distributing −n-n to both terms inside the parentheses.
n2−n(1−n)=n2−n+n2n^2 - n(1-n) =n^2 - n + n^2

STEP 7

Now, simplify the expression.
7n2−n(1−n)=7n2−n+n2=n2−n7n^2 - n(1-n) =7n^2 - n + n^2 =n^2 - nThe correct answer for problem3 is A. n2−nn^2 - n.

STEP 8

For problem4, we need to solve the inequality t3−1<t\frac{t}{3}-1 < t. This can be done by subtracting t3\frac{t}{3} from both sides.
t3−1−t3<t−t3\frac{t}{3}-1 - \frac{t}{3} < t - \frac{t}{3}

STEP 9

Now, simplify the inequality.
t3−−t3<t−t3=−<2t3\frac{t}{3}- - \frac{t}{3} < t - \frac{t}{3} = - < \frac{2t}{3}

STEP 10

To isolate tt, we multiply both sides of the inequality by 32\frac{3}{2}.
−⋅32<2t3⋅32- \cdot \frac{3}{2} < \frac{2t}{3} \cdot \frac{3}{2}

STEP 11

Now, calculate the value of tt.
−⋅3<t3⋅3=−3<t- \cdot \frac{3}{} < \frac{t}{3} \cdot \frac{3}{} = -\frac{3}{} < tThe correct answer for problem4 is C. −3<t\frac{-3}{} < t.

STEP 12

For problem5, we need to solve the inequality a+>2a−2a+ > \frac{}{2}a-2. This can be done by subtracting 2a\frac{}{2}a from both sides.
a+−2a>2a−2−2aa+ - \frac{}{2}a > \frac{}{2}a-2 - \frac{}{2}a

STEP 13

Now, simplify the inequality.
a+−2a>2a−2−2a=2a+>−2a+ - \frac{}{2}a > \frac{}{2}a-2 - \frac{}{2}a = \frac{}{2}a + > -2

STEP 14

To isolate aa, we subtract from both sides of the inequality.
2a+−>−2−\frac{}{2}a + - > -2 -

STEP 15

Now, calculate the value of aa.
2a>−3\frac{}{2}a > -3

STEP 16

To get aa by itself, we multiply both sides of the inequality by2.
2⋅2a>2⋅−32 \cdot \frac{}{2}a >2 \cdot -3

STEP 17

Now, calculate the value of aa.
a>−6a > -6The correct answer for problem5 is A. a>−6a > -6.

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