Ask question

Ask question

All categories

GenaCL600 [577]

3 years ago

11

an average ant is 1/4 of an inch long. an average aphid is 3/32 of an inch long. how many times longer is the ant than the aphid

1 answer:

Nataly_w [17]3 years ago

3 0

First find the common denominator. In this case, you would want to use 32. So, 1/4 = 8/32. So, an ant is 8/32 of an inch and an aphid is 3/32 of an inch. Now to find the difference, you subract. 8/32 - 3/32 = 5/32. So the answer is, the ant is 5/32 of an inch longer.

You might be interested in

Statistical Question

nadezda [96]

Answer:

c

Step-by-step explanation:

5 0

3 years ago

Drag the tiles to the boxes to form correct pairs.<br> 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.

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

<em>Solving the expression</em>

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

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

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

$10t+5-7t+28$

$3t+33$

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

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

<em>Solving the expression</em>

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

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

$9t-12-2t-10$

$7t-22$

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

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

<em>Solving the expression</em>

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

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

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

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

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

As

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

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

<em>Solving the expression</em>

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

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

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

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

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

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

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

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

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

#learnwithBrainly

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=4x%20-%20%20%7Bx%7D%5E%7B3%7D%20" id="TexFormula1" title="4x - {x}^{3} " alt="4x - {x}^{3} "

nydimaria [60]

X(2+x)(2-x) I think that’s correct

3 0

3 years ago

Read 2 more answers

Jaime graphed the quadratic y=-2x^2-12x+3. The equation of the line of symmetry of the function is x=-3 What is the vertex of th

miv72 [106K]

Vertex is (-3 , 21 ) you can put <span>y=-2x^2-12x+3 in google and you will see the function</span>

3 0

3 years ago

Read 2 more answers

How can you write 23/3 as a mixed number?

Illusion [34]

23/3 = 7 2/3 <======

3 0

3 years ago