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3 years ago

Dora spent $45 and has $32 left. Which equation can be used to calculate how much money Dora had at the beginning?

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

7 0

32+45 can be used to figure out the starting amount

vladimir2022 [97]3 years ago

5 0

$45 + $32 I believe -?
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3 years ago

Solve 2(x-2) = 3(x+2)-8

IgorC [24]

Answer:

x = -2

Step-by-step explanation:

2(x-2) = 3(x+2)-8

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3 years ago

the test scores for 10 students were 61,67,81,83,87,88,89,90,98, and 100. which frequency table represents this data set?

lana66690 [7]

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below only shows up once

Step-by-step explanation:

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3 years ago

Prove that √2 +√5 is irrational

Sindrei [870]

We have to prove that $\sqrt{2}+\sqrt{5}$ is irrational. We can prove this statement by contradiction.

Let us assume that $\sqrt{2}+\sqrt{5}$ is a rational number. Therefore, we can express:

$a=\sqrt{2}+\sqrt{5}$

Let us represent this equation as:

$a-\sqrt{2}=\sqrt{5}$

Upon squaring both the sides:

$(a-\sqrt{2})^{2}=(\sqrt{5})^{2}\\a^{2}+2-2\sqrt{2}a=5\\a^{2}-2\sqrt{2}a=3\\\sqrt{2}=\frac{a^{2}-3}{2a}$

Since a has been assumed to be rational, therefore, $\frac{a^{2}-3}{2a}$ must as well be rational.

But we know that $\sqrt{2}$ is irrational, therefore, from equation $\sqrt{2}=\frac{a^{2}-3}{2a}$ the expression $\frac{a^{2}-3}{2a}$ must be irrational, which contradicts with our claim.

Therefore, by contradiction, $\sqrt{2}+\sqrt{5}$ is irrational.

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3 years ago

I don't understand that question, please help<br> Thank u

chubhunter [2.5K]

Answer:

.I didn't understand it either sorry

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3 years ago