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3 years ago

8

Daniel's graduation will cost $9 for every 3 people who attend. At most how many attendees can there be if Daniel budgets a tota

l of $18 for his graduation?

2 answers:

ella [17]3 years ago

7 0

Answer:

6

Step-by-step explanation:

6 because 9*2=18 and 3*2=6

bearhunter [10]3 years ago

5 0

Answer: 6

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Can you please help me with this there is a pic on the top thx

Svetach [21]

It would be 1000 x 6
so the answer would be 6000.
hope this helped but plz help me out and put this answer as brainliest plz

6 0

4 years ago

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Which graph shows the solution to the inequality −2(2x+3)≤26?

krok68 [10]

Answer:

x≥−8

Step-by-step explanation:

Step 1: Simplify both sides of the inequality.

−4x−6≤26

Step 2: Add 6 to both sides.

−4x−6+6≤26+6

−4x≤32

Step 3: Divide both sides by -4.

−4x/−4 ≤ 32/−4

x≥−8

5 0

3 years ago

if two pyramids are similar and the ratio between the lengths of their edges is 2;7 what is the ratio of their volumes? ...?

jok3333 [9.3K]

<span>The answer is 8:343.

</span>We can use Galileo's square cube law to calculate the ratio between two similar pyramids. The law is used to describe the change of the area or the volume of the shape when their dimensions increase or decrease:
$\frac{V_1}{V_2} =( \frac{l_1}{l_2} )^{3} \\$ <span>
</span><span><span> <span> <span> <span> <span> <span> <span> <span> <span> <span> <span> </span></span></span></span></span></span></span></span></span> </span></span> </span>
V₁ and V₂ - volumes of pyramids,
l₁ and <span>l</span>₂ - the edges of pyramids.
<span>
</span> $\frac{V_1}{V_2} =( \frac{l_1}{l_2} )^{3} \\ \frac{V_1}{V_2} =( \frac{2}{7} )^{3} \\ \frac{V_1}{V_2} = \frac{2^{3}}{7^{3}} \\ \frac{V_1}{V_2} = \frac{8}{343} \\$ <span>
</span>

3 0

3 years ago

Read 2 more answers

If a series of rigid transformations maps ∠F onto ∠C where ∠F is congruent to ∠C, then which of the following statements is true

lesantik [10]

Answer:

the correct answer is A good luck

Step-by-step explanation:

8 0

3 years ago

Factor out the GCF from the terms of the polynomial –4y5 + 6y3 + 8y2 – 2y. A. –2y4 + 3y2 + 4y – 1 B. y(–4y4 + 6y2 + 8y – 2) C. 2

andreyandreev [35.5K]

Answer:

Option D) $-2y(2y^4-3y^2-4y+1)$ is correct

Therefore $-4y^5+6y^3+8y^2-2y=-2y(2y^4-3y^2-4y+1)$

Step-by-step explanation:

Given polynomial is $-4y^5+6y^3+8y^2-2y$

To factorise the given polynomial by taking out the common terms of the given polynomial :

$-4y^5+6y^3+8y^2-2y$
$=2y(-2y^4+3y^2+4y-1)$ ( here 2y is GCF common term so taking outside the terms of the polynomial )
$=-2y(2y^4-3y^2-4y+1)$ ( now taking (-) outside )

Therefore $-4y^5+6y^3+8y^2-2y=-2y(2y^4-3y^2-4y+1)$

Option D) $-2y(2y^4-3y^2-4y+1)$ is correct

8 0

4 years ago