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3 years ago

8

Simplify. |–10 – 3| =

1 answer:

Damm [24]3 years ago

6 0

I think its -13. Because, you have to add -10 and -3 then, the answers is -13.

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What is the probability that the first to two checkers he pulls from the box at random will be two red checkers?

UkoKoshka [18]

It is 1 out of 24 chance do draw a red a first time.

Double the 24 for the second time to get the correct answer.

1/48 percent change of getting 2 red checkers in a row.

If you need anymore help, let me know.

Brainliest answer is always appreciated!

8 0

3 years ago

Which of the following is true about the expression

antoniya [11.8K]

It’s b, square root of 3 is irrational, 2 is rational, it creates a irrational number

8 0

3 years ago

A train goes twice as fast downhill as it can go uphill, and 2/3 feet as fast uphill as it can go on level ground. If it goes 12

Yuri [45]

Answer:

0.5hrs

Step-by-step explanation:

Let the speed downhill be $x$ ,

Speed uphill is $0.5x$ . We are told than speed uphill is $\frac{2}{3}$ the speed on flat ground. Therefore the speed on level ground will be $\frac{0.5x}{2/3}=0.75x$ .

From this info, we can obtain the actual speed, $v_g$ on level ground as:

$x=120m/h\\0.75x=v_g\\\\\therefore v_g=\frac{0.75x \times 120m/h}{x}\\v_g=90m/h$

speed, distance and time have the relation $s=d/t$ so to obtain time to cover 45miles:

$t=d/s\\=45/90\\=0.5h$

Hence it takes 0.5hrs to cover 45miles on flat ground.

4 0

3 years ago

Choose all the functions that are NOT linear.

Triss [41]

Answer:
B

Step:
The general equation for linear is y=mx+b

So all is linear except B

8 0

2 years ago

Write the standard form equation of the ellipse shown in the graph, and identify the foci.

vlada-n [284]

Answer option A

From the given graph is a Vertical ellipse

Center of ellipse = (-2,-3)

Vertices are (-2,3) and (-2,-9)

Co vertices are (-6,-3) and (2,-3)

The distance between center and vertices = 6, so a= 6

The distance between center and covertices = 4 , so b= 4

The general equation of vertical ellipse is

$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2}=1$

(h,k) is the center

we know center is (-2,-3)

h= -2, k = -3 , a= 6 and b = 4

The standard equation becomes

$\frac{(x+2)^2}{4^2} + \frac{(y+3)^2}{6^2}=1$

$\frac{(x+2)^2}{16} + \frac{(y+3)^2}{36}=1$

Foci are (h,k+c) and (h,k-c)

$c=\sqrt{a^2-b^2}$

Plug in the a=6 and b=4

$c=\sqrt{6^2-4^2}$

$c=\sqrt{20}$

$c=2\sqrt{5}$ , we know h=-2 and k=-3

Foci are $(-2,-3+2\sqrt{5})$ and $(-2,-3-2\sqrt{5})$

Option A is correct

6 0

3 years ago