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3 years ago

So i have a pic i need 1a plz i am giving a bunch of my points plz give it real asap

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

6 0

Answer:

3x - (15 x 3) = 90

3x - 45 = 90

Explanation:

there are 3 boxes. We don't know how many are in them yet, so we write this as 3x. However, we know that we give away 15 to 3 different teachers. 15 x 3 = 45

Therefore the first part is 3x - 45

We also know that in the end we have 90 left.

So our final equation is:

3x - 45 = 90

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ethan made a rectangular prism that has a volume of 16 cubic inches. What is one set of possible dimensions for ethans rectangul

lions [1.4K]

Answer:

0.5 inches by 0.5 inches by 64 inches

Step-by-step explanation:

Any three positive dimensions that have a product of 16 will be a set of possible dimensions. There are an infinite number of them. One such is listed above.

(If you have a list of answer choices, you can multiply them out to see which have a product of 16. We cannot guess what your choices might be.)

4 0

3 years ago

Indicate the answer choice that best completes the statement or answers the question.

Tema [17]

Answer:

Step-by-step explanation:

m<2:

180 - 60 - 48 = 72

Vertical angle thm:

m<1 = 48

m<3:

180 - 62 - 48 = 70

6 0

3 years ago

Find the z-score such as 88% of the area under the normal curve is to the left of the z-score

jok3333 [9.3K]

If you have a calculator with statistical functions, that's the way to go.

On my TI-83, I typed in invNorm(0.88) and got the result z = 1.17.

88% of the area under the normal curve is to the left of z = 1.17.

6 0

3 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$