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2 years ago

9 less than the product is 7 and x is 75..

Mathematics

1 answer:

makvit [3.9K]2 years ago

7 0

The answer is 7x - 9 = 75. Please mark me brainiest

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The transformation (x,y) -> (x + 4,y – 3) is performed on the line segment AB.

exis [7]

It’s either d or A but I’m sure it’s A

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3 years ago

Heeeelp just need 1 and 3

Harlamova29_29 [7]

Answer: I think the answer is. x=y*z I dont know what the verbles are but thats what I got. ANd can you zoom in a bit?

Step-by-step explanation:

4 0

2 years ago

PLEASE HELP IM STRUGGLING!!!

Mnenie [13.5K]

Answer:

10. 0.99
11. -60
12. 5.5

Step-by-step explanation:

Use the slope formula:

m = (y2 - y1)/(x2 - x1)

10.

m = (6.24 - 3.27)/(5 - 2) = 2.97/3 = 0.99

11.

m = (240 - 360)/(3 - 1) = -120/2 = -60

12.

m = (8.84 - 6.09)/(7 - 2) = 2.75/5 = 5.5

6 0

3 years ago

Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.

leonid [27]

Answer-

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

$12x-5y+14=0 \\ 12x-5y-14=0$

Solution-

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1$

$\Rightarrow 12h-5k-1=\pm 13$

$\Rightarrow 12h-5k=\pm 14$

$\Rightarrow 12h-5k=14,\ 12h-5k=-14$

$\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0$

$\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0$

Two equations are formed because one will be upper from the the given line and other will be below it.

4 0

2 years ago

Answers:

ryzh [129]

Answers:

☆☆☆☆☆☆☆☆aTriangle ZYX
bTriangle YZX
cTriangle XZY
dTriangle XYZ

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3 years ago