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zhannawk [14.2K]

3 years ago

9

Please help me please

1 answer:

Natasha_Volkova [10]3 years ago

8 0

I THINK ITS C HAVE A GOOD DAY

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−1/3−(−1/2)

Radda [10]

Hi ;-)

$-\frac{1}{3}-(-\frac{1}{2})=-\frac{1}{3}+\frac{1}{2}=-\frac{2}{6}+\frac{3}{6}=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\\\\3\frac{1}{3}-5=-(5-3\frac{1}{3})=-(4\frac{3}{3}-3\frac{1}{3})=-1\frac{2}{3}\\\\-1\frac{4}{5}-(-2\frac{7}{8})=-1\frac{4}{5}+2\frac{7}{8}=-1\frac{32}{40}+2\frac{35}{40}=2\frac{35}{40}-1\frac{32}{40}=1\frac{3}{40}\\\\-3\frac{3}{8}-\frac{7}{8}=-(3\frac{3}{8}+\frac{7}{8})=-3\frac{10}{8}=-3\frac{5}{4}=-4\frac{1}{4}\\\\4\frac{3}{4}-(-1\frac{1}{12})=4\frac{9}{12}+1\frac{1}{12}=5\frac{10}{12}$

6 0

3 years ago

What is the image of A(8,2) under R90 ?<br> A. (-2,8)<br> B. (2,8)<br> C.( 8,-2)<br> D.(-8,2)

swat32

Answer: A. (-2,8)

Step-by-step explanation:

We know that the rule for rotation of $R_{90}$ is given by :-

$(x,y)\rightarrow(-y,x)$ , where (x,y) is the point of pre-image and (-y,x) is a point of image.

The given point of pre-image : A(8,2)

Then after rotation of $R_{90}$ , we have the image point as :-

$(8,2)\rightarrow(-2,8)$

Hence, the image point of A(8,2) is (-2,8) .

5 0

3 years ago

Whats the distance between (-3,4) and (5,1)?

guajiro [1.7K]

Answer:

Step-by-step explanation:

(-3,4) and (5,1)

(-3,4) is in quadrant 2 and (5,1) is in quadrant 1. So think of that.

(x+8, y-3)

8 units to the right and 3 units down

3 0

3 years ago

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What is the distance between two points with coordinates at (8, 40) and (20, 5)?

lora16 [44]

Answer:

b hope it helps

Step-by-step explanation:

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3 years ago

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Given the function f(x)= x^4+6x^3-x^2-30x+4

Romashka-Z-Leto [24]

The correct answers are a,b,e,f

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4 years ago