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3 years ago

Multiply.

Mathematics

2 answers:

shtirl [24]3 years ago

7 0

When you multiply a negative and a positive you will get a negative answer so 3*7 is 21 so -3 * 7 is -21

solong [7]3 years ago

6 0

The answer is -21 because take 3 times 7 which is 21 and sense there's a negative and positive you would make 21 negative.

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A+ -12= 45 6th grade

Dmitry_Shevchenko [17]

A = 57
45 + 12 = 57
57 - (-12) = 45

8 0

3 years ago

Which statement is an example of the symmetric property of congruence?

BabaBlast [244]

The <u>correct answer</u> is:

$\text{D) If }EFG \cong HJK, \text{ then } HJK \cong EFG.$

Explanation:

The symmetric property states that if two quantities are equal (or congruent), you can "flip" them around the equals sign (or congruence sign).

This means since EFG is congruent to HJK, then HJK is congruent to EFG.

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3 years ago

A container is 24 m long, 9 m wide and 8 m high.

GuDViN [60]

Answer:

900,000 books.

Step-by-step explanation:

The volume of the container = 24 * 9 * 8

= 1728 m^3

The volume of each book = 20*16*6 = 1920 cm^3

There are 100*100*100 = 1,000,000 cm^3 in 1 m^3.

So the number of books the container can hold

= (1728 * 1000000) / 1920

= 900,000.

5 0

3 years ago

3=4x+8y i need help

Daniel [21]

The answer would be if you were solving for y. Y= 1/2x+3/8

6 0

3 years ago

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

3) To test it, just integrate:

$\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C$

5 0

3 years ago