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2 years ago

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Sean can spend at most $15.00 on snacks. He has ailready spent s5.00. which inequality could be solved to determine how much mon

ey Sean has left to spend on snacks?

1 answer:

liq [111]2 years ago

7 0

Answer:

https://cougar.collegiate-va.org/cfoster/Mrs._Fosters_Classes/Algebra_1_files/0437_001.pdf

Step-by-step explanation: go to the 3rd page of link

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Chantel drew a picture of her dog on a piece of paper that is 12 centimeters long. She used a copy machine to enlarge her drawin

Georgia [21]

Answer:

a(n)=1.15[a(n-1)]

Step-by-step explanation:

we know that

$115\%=115/100=1.15$

Let

a0 -----> the length of the original copy

<em>The first copy is equal to</em>

a1=1.15(a0)

<em>The second copy is </em>

a2=1.15[1.15(a0)] or a2=1.15[a1]

<em>The third copy is</em>

a3=1.15{1.15[1.15(a0)]} or a3=1.15[a2]

therefore

A recursive formula will be

a(n)=1.15[a(n-1)]

3 0

3 years ago

Ahmad is choosing a password to access his teacher’s web page. He must choose a capital letter and three nonrepeating digits fro

algol [13]

Answer:

0a23,02A3

Step-by-step explanation:

I think that this is the right answer Im sorry if it is wrong

7 0

2 years ago

Read 2 more answers

The sum of three numbers is 97. The third number is 2 times the second. The second number is 9 less than the first. What are the

inn [45]

Answer:

The numbers are 26.5, 53 and 17.5.

Step-by-step explanation:

97=x+(x-9)+2x

97+9=4x

106/4=4x/4

x= 26.5

x-9 = 17.5

2x= 53

8 0

3 years ago

Do the sides 4, 5, and 6 form a Pythagorean Triple?

NemiM [27]

Answer:

They do not form a Pythagorean Triple.

If it was a Triple, 4^2+5^2 should equal 6^2.

4^2+5^2=16+25=41

6^2=36

41≠26

It doesn't work, therefore it is not a Pythagorean Triple.

3 0

2 years ago

Read 2 more answers

Solve each inequality, and then drag the correct solution graph to the inequality.

Nesterboy [21]

The correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

(NOTE: The graphs are labelled A, B and C from left to right)

For the first inequality,

$4(9x-18)>3(8x+12)$

First, clear the brackets,

$36x-72>24x+36$

Then, collect like terms

$36x-24x>36+72\\12x >108$

Now divide both sides by 12

$\frac{12x}{12} > \frac{108}{12}$

∴ $x > 9$

For the second inequality

$-\frac{1}{3}(12x+6) \geq -2x +14$

First, clear the fraction by multiplying both sides by 3

$3 \times[-\frac{1}{3}(12x+6)] \geq3 \times( -2x +14)$

$-1(12x+6) \geq -6x +42$

Now, open the bracket

$-12x-6 \geq -6x +42$

Collect like terms

$-6 -42\geq -6x +12x$

$-48\geq 6x$

Divide both sides by 6

$\frac{-48}{6} \geq \frac{6x}{6}$

$-8\geq x$

∴ $x\leq -8$

For the third inequality,

$1.6(x+8)\geq 38.4$

First, clear the brackets

$1.6x + 12.8\geq 38.4$

Collect likes terms

$1.6x \geq 38.4-12.8$

$1.6x \geq 25.6$

Divide both sides by 1.6

$\frac{1.6x}{1.6}\geq \frac{25.6}{1.6}$

∴ $x \geq 16$

Let the graphs be A, B and C from left to right

The first graph (A) shows $x\leq -8$ and this matches the 2nd inequality

The second graph (B) shows $x \geq 16$ and this matches the 3rd inequality

The third graph (C) shows $x > 9$ and this matches the 1st inequality

Hence, the correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

Learn more here: brainly.com/question/17448505

8 0

2 years ago