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3 years ago

5

A bag contains 7 red cubes, 3 yellow cubes, 2 green cubes, and 4 blue cubes, one cube is selectef at random from the bag. What i

s the probability that the seleceted cube will not be blue.

1 answer:

stich3 [128]3 years ago

5 0

The answer is 12/16 because there’s 4 blue cubes and 16 cubes total .

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A lighthouse is 90 feet above sea level. The angle of depression from the top of the lighthouse to the center of the rock is 35

Marysya12 [62]

Notice the picture below

so.... the angle of depression, is above in dashed lines.... however, the angle of depression, is identical to the angle of elevation from the water, since, both angles are "alternate interior" angles

now, using the 45-45-90 rule, then we know the swimmer is 90 feet from the lighthouse, then just use the tangent ratio to get "x"

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3 years ago

6 cousins inherited a total of 58 of their great aunt's assets which they divided evenly what fraction of their great aunt's ass

mina [271]

Answer:

9 1/9 per cousin

Step-by-step explanation:

if the aunt has 58 items to be distributed to 6 cousins, we would set a fraction of 58/6. performing that is 9 with 4/6 left over.

since there is 6 cousins, there will be 2/3 divided by 6 = 2/18, or 1/9

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2 years ago

The leg of a right triangle is 3 units and the hypotenuse is 11 units. What is the length, in units, of the other leg of the tri

son4ous [18]

Answer:

D

Step-by-step explanation:

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3 years ago

Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

Ede4ka [16]

Answer:

Part a) The quadratic function is $4x^{2} +20x-39=0$

Part b) The value of x is $1.5\ in$

Part c) The photo and frame together are $7\ in$ wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

$(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0$

Part b) What is the value of x?

Solve the quadratic equation $4x^{2} +20x-39=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem

we have

$4x^{2} +20x-39=0$

so

$a=4\\b=20\\c=-39$

substitute in the formula

$x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}$

$x=\frac{-20(+/-)\sqrt{1,024}} {8}$

$x=\frac{-20(+/-)32} {8}$

$x=\frac{-20(+)32} {8}=1.5\ in$ -----> the solution

$x=\frac{-20(-)32} {8}=-6.5\ in$

Part c) How wide are the photo and frame together?

$(4+2x)=4+2(1.5)=7\ in$

5 0

3 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

3 0

3 years ago