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Blababa [14]
3 years ago
5

Please help finding area

Mathematics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

so what's 10 x 15?

that's what it's mainly asking you

10 x 15

answer: is 150

- that's what I think it is..

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Hello, may someone plz help me
fenix001 [56]

Answer:

Step-by-step explanation:

this is not high school- but uhm

1) KAT

2)TAK

3) SAT

i think thats the answer for the first section..

4 0
3 years ago
Y+7=-2(x-1)<br><br><br> hurry plz
zhannawk [14.2K]

Answer:

y=-2x-5

Step-by-step explanation:

First of all, you have to distribute -2 to (x-1) to get to this equation y+7=-2x+2.  Then, you subtract 7 on both sides to get the slope-intercept of y = -2x - 5.

8 0
3 years ago
A truck has a force of 2000 newtons and is moving 10 miles per hour. How much mass does the truck have?
worty [1.4K]

Answer:

Force= Mass×Accelaraion

F=m×a

But as The unit of accelaration is given miles per hour and the SI is meters per second sqaure we have to convert 10mph to m/s²

Thus, we have 10mph = 4.47 meters per second square. (i converted using scientific calculator)

So now we have,

2000N= m×4.47m/s²

= 2000/4.47m/s²=m

= 447.42

Thus the mass of the object is 447.42 (i am not sure of units)

7 0
2 years ago
Read 2 more answers
Help Are the triangles similar? If so, determine the scale factor of ΔPQR to ΔEFD.
alex41 [277]
The answer is B. 
When you set up a ratio of all the corresponding sides, the ratios all equal 2:1
8 0
3 years ago
For how many real values of x is <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B120-%5Csqrt%7Bx%7D%7D%20" id="TexFormula1" title
leva [86]

A good place to start is to set \sqrt{x} to y. That would mean we are looking for \sqrt{120-y} to be an integer. Clearly, y\leq 120, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since \sqrt{x} is a radical, it only outputs values from [0,\infty], which means y is on the closed interval: [0,120].

With that, we don't really have to consider y anymore, since we know the interval that \sqrt{x} is on.

Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval [0,120], which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

\sqrt{120-\sqrt{x}}=k \implies \\ \sqrt{x}=k^2-120 \implies\\ x=(k^2-120)^2

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.

5 0
3 years ago
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