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Mekhanik [1.2K]

2 years ago

5

12x0.08 help me plss!!!

2 answers:

mel-nik [20]2 years ago

6 0

Answer:

0.96

Step-by-step explanation:

Sphinxa [80]2 years ago

5 0

Answer:

If the x is a times sign, then the answer is: 0.96

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Convert 35 mi/h to km/h

notsponge [240]

<span>Convert 35 mi/h to km/h

</span> $35 \dfrac{mi}{h}\to \dfrac{km}{h}\qquad \qquad 1 mi= 1,60934 \\ \\ \\ 35 \dfrac{mi}{h}* \dfrac{1,60934\ km}{1\ mi}= \boxed{ 56,3269\dfrac{km}{h} }$ <span>
</span>

3 0

3 years ago

1. Match the polynomial identity to the correct answer.

Katen [24]

Answer:

$a: (a-b)(a+b);b: (a-b)(a^{2}+ab+b^{2}); c: (a^{2} +2ab+b^{2}); d: (a+b)(a^{2}- ab+b^{2})$

Step-by-step explanation:

Given the expressions all that is required is to expand them (using the distributive property) to see which answers they match up to:

For a: $(a-b)(a+b)=a^{2} +ab-ab-b^{2} =a^{2}-b^2$

For b: $(a-b)(a^2+ab+b^2)=a^3+a^2b+ab^2-a^2b-ab^2-b^3=a^3-b^3$

For c calculate the square of a sum: $(a+b)^2=a^2+ab+ab+b^2$

(add like terms) c: $a^2+2ab+b^2$

For d: $(a+b)(a^2-ab+b^2)=a^3-a^2b+ab^2+a^2b-ab^2+b^3$ $=a^3+b^3$

5 0

3 years ago

Which graph shows a linear function?

dimulka [17.4K]

Answer:

I this that its the third graph

8 0

2 years ago

I need help with this question

Novay_Z [31]

Answer:

$\frac{\sqrt{3} - 1}{2\sqrt{2}}$

$\frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

$- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Step-by-step explanation:

Given $\frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6}$

(A) $sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4} + \frac{\pi}{6})$

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2}$

$$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

(B) Cos(A + B) = CosAcosB - SinASinB

$cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}})$

$\implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2}$

$\implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6})$

$$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

(C) Tan(A + B) = $\frac{Sin(A +B)}{Cos(A + B)}$

From the above obtained values this can be calculated and the value is $- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .

3 0

3 years ago

If the table coordinates for triangle XYZ and it's images using two transformations use 1 grid on both axes to represent 1 unit

Svetlanka [38]

Answer:

I'm not good at math

Step-by-step explanation:

7 0

3 years ago