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3 years ago

If a term is multiplied by zero, the result product will always be

1 answer:

4 0

Hi there!

If any term, no matter what the number multiplied by zero will always turn out to be zero. I can give you an example of why this is correct.

We'll use apples. :P

If we have four apples multiplied by zero, you'd have zero apple because there's NO groups of apples to use.

Two times three is six because there are two groups of three, or three groups of two. But since earlier we said that there were zero groups of apples, there is zero groups of four so that cancels out to make none.

Hope this helps! :D

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Jack used 40 balloons to decorate a room for a party. Of the balloons he used, 15% were red balloons.

ExtremeBDS [4]

15% of 40 = 6 so I’m guessing it would be 6

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3 years ago

Which ones are equivalent tell me all that apply:)

Snezhnost [94]

a) second and last one : 32x - 24 and 8*4x - 8*3

b) first and third one : 22y and 7y + 15y

pls mark brainliest!!

3 0

3 years ago

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Can someone help me and explain? I will mark brainlest. ♡

Sedbober [7]

Answer:

$p'(4) = -3$

$q'(8) = \frac{1}{4}\\$

Step-by-step explanation:

For p'(4):

$p(x) = f(x)g(x) \\ p'(x) = \frac{d}{dx}(f(x)g(x)) \\ p'(x) = f'(x)g(x) +f(x)g'(x)$

$p'(4) = f'(4)g(4) + f(4)g'(4) \\ p'(4) = (-1)(3) +(7)(0) \\ p'(4) = -3$

For q'(8):

$q(x) = \frac{f(x)}{g(x)} \\ q'(x)= \frac{d}{dx}(\frac{f(x)}{g(x)}) \\ q'(x) = \frac{f'(x)g(x) -f(x)g'(x)}{{g(x)}^2}$

$q'(8) = \frac{f'(8)g(8) -f(8)g'(8)}{{g(8)}^2} \\ q'(8) = \frac{(2)(2) -(6)(\frac{1}{2})}{{2}^2} \\ q'(8) = \frac{4 -3}{4} \\ q'(8) = \frac{1}{4}$

4 0

3 years ago

Read 2 more answers

The length of each side of an equilateral triangle is increased by 20%, resulting in triangle ABC. If the length of each side of

kompoz [17]

Answer: Area of ΔABC is 2.25x the area of ΔDEF.

Step-by-step explanation: Because equilateral triangle has 3 equal sides, area is calculated as

$A=\frac{\sqrt{3} }{4} a^{2}$

with a as side of the triangle.

Triangle ABC is 20% bigger than the original, which means its side (a₁) measures, compared to the original:

a₁ = 1.2a

Then, its area is

$A_{1}=\frac{\sqrt{3} }{4}(1.2a)^{2}$

$A_{1}=\frac{\sqrt{3} }{4}1.44a^{2}$

Triangle DEF is 20% smaller than the original, which means its side is:

a₂ = 0.8a

So, area is

$A_{2}=\frac{\sqrt{3} }{4} (0.8a)^{2}$

$A_{2}=\frac{\sqrt{3} }{4} 0.64a^{2}$

Now, comparing areas:

$\frac{A_{1}}{A_{2}}= (\frac{\sqrt{3}.1.44a^{2} }{4})(\frac{4}{\sqrt{3}.0.64a^{2} } )$

$\frac{A_{1}}{A_{2}} =$ 2.25

The area of ΔABC is 2.25x greater than the area of ΔDEF.

7 0

2 years ago

What is the sine value of 7 pi over 6? negative square root 3 over 2 square root 3 over 2 negative 1 over 2 1 over 2?

djverab [1.8K]

First thing to do is to change the radians to degrees so it's easier to determine our angle and where it lies in the coordinate plane. $\frac{7 \pi }{6} * \frac{180}{ \pi }=210$ . If we sweep out a 210 degree angle, we end up in the third quadrant, with a 30 degree angle. In this quadrant, x and y are both negative, but the hypotenuse, no matter where it is, will never ever be negative. So the side across from the 30 degree reference angle is -1, and the hypotenuse is 2, so the sine of this angle, opposite over hypotenuse, is -1/2

6 0

3 years ago

Read 2 more answers