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kogti [31]
3 years ago
12

HELP!!! I will give brainliest, 25 points!!

Mathematics
1 answer:
kirza4 [7]3 years ago
6 0

Cards = x

Bouquets = y

They spend 2 on each card so 2x and 3.50 for each bouquet so 3.50y add those together to get the total they can spend 2x + 3.50y = 360

Then add the sales together: 6x + 8y = 900

Answer: D. 2x + 3.50y = 360; 6x + 8y = 900

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Help answer this plz
Sphinxa [80]
X=-1 is the answer. Hope that helps!
8 0
4 years ago
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Allen is running a marathon. He would like to finish the race in under 5.5 hours. He has already run 3 hours. The inequality rep
AlexFokin [52]

Answer:

x < 2.5

Step-by-step explanation:

Allen has already run 3 hours and will run x amount more to finish. This is 3+x. He must finish under or less than 5.5 hours. So 3 + x < 5.5. You solve and find x < 2.5.

7 0
3 years ago
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WILL GIVE BRAINLIEST&amp; 15 PTS.
kicyunya [14]

Answer:

The given fraction \frac{x^3-x^2}{x^3} reduces to  \frac{x-1}{x}

Step-by-step explanation:

Consider the given fraction \frac{x^3-x^2}{x^3}

We have to reduce the fraction to the lowest terms.

Consider numerator x^3-x^2

We can take x² common from both the term,

Thus, numerator can be written as x^2(x-1)

Given expression can be rewritten as ,

\frac{x^3-x^2}{x^3}=\frac{x^2(x-1)}{x^3}

We can now cancel x^2 from both numerator and denominator,

\Rightarrow \frac{x^2(x-1)}{x^3}=\frac{x^2(x-1)}{x^2 \cdot x}

\Rightarrow \frac{(x-1)}{x^2 \cdot x}=\frac{x-1}{x}

Thus, the given fraction \frac{x^3-x^2}{x^3} reduces to  \frac{x-1}{x}

6 0
3 years ago
Read 2 more answers
Write the equation of the line that passes through the point (7,-2) and has a slope of
konstantin123 [22]
I only know how to do a, so i hope i can help a little bit.
y=mx+b
-2=-3(7)+b
-2=-21+b
b=19
so your equation is y=-3x+19
i hope this helps!!
6 0
3 years ago
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Prove that root 7 is irrational by the method of contradiction
Alchen [17]

Let assume that \sqrt7 is a rational number. Therefore it can be expressed as a fraction \dfrac{a}{b} wherea,b\in\mathbb{Z} and \text{gcd}(a,b)=1.

\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2

This means that a^2 is divisible by 7, and therefore also a is divisible by 7.

So, a=7k where k\in\mathbb{Z}

(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2

Analogically to a^2=7b^2 ------- b^2 is divisible by 7 and therefore so is b.

But if both numbers a and b are divisible by 7, then \text{gcd}(a,b)=7 which contradicts our earlier assumption that \text{gcd}(a,b)=1.

Therefore \sqrt7 is an irrational number.

8 0
4 years ago
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