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1 year ago

Question 1 of 10 Solve 10x+16 ≥ 6x+ 20. O A. x29 OB. x≤ 1 OC. x2 1 OD. x≤ 9

Mathematics

1 answer:

Alexxandr [17]1 year ago

6 0

Answer: I think B. Explain:

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Drag the tiles to the boxes to form correct pairs. Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

Solving the expression

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

Solving the expression

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

Solving the expression

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

Solving the expression

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

Learn more about algebraic expression from brainly.com/question/11336599

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5 0

3 years ago

Read 2 more answers

Harry ate 5/8 of a sandwich

11111nata11111 [884]

Well do you know a common factor between 8, 5, and 12, if so it is super simple to find your answer. :)

4 0

3 years ago

Read 2 more answers

Pls can someone help me <img src="https://tex.z-dn.net/?f=%20log10%20%5Csqrt%7B36%7D%20%20%2B%20%20%20%20log10%20%20%5Csqrt%7

Firlakuza [10]

Answer:

$= log_{10}(\frac{6\sqrt{2} }{\sqrt{7} } )$

Step-by-step explanation:

Given the expression $log10\sqrt{36} + log10\sqrt{2} - log10\sqrt{7} \\$

According to the law of logarithm;

loga + log b = log(ab) and log a - log b = log(a/b)

The given expression becomes;

$log10\sqrt{36} + log10\sqrt{2} - log10\sqrt{7} \\= log_{10}(\frac{\sqrt{36} \times \sqrt{2} }{\sqrt{7} })\\= log_{10}(\frac{6\sqrt{2} }{\sqrt{7} } )$

6 0

2 years ago

Factor to find the zeros of the function:

enyata [817]

Step-by-step explanation:

y = 8x² + 8x - 96

y = 8(x² + x - 12)

y = 8(x + 4)(x - 3)

When y = 0, 8(x + 4)(x - 3) = 0.

=> x = -4 or x = 3. (B)

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2 years ago

Which best defines a circle?

Paraphin [41]

I believe the correct answer is A,

6 0

3 years ago