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3 years ago

What operation is being done to the variable in the equation -5m= -40?

Mathematics

1 answer:

VLD [36.1K]3 years ago

6 0

The answer would be A) It's being multiplied by -5

This is because whenever a variable is touching a number without anything in between it, it is always multiplication.

I hope this helps!

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The base of the pyramid is a rectangle that is not a

Harrizon [31]

An axis of symmetry passes through the vertex and is perpendicular to the base. Then the correct options are B, C, and E.

<h3>What is a rectangular pyramid?</h3>

Take a rectangle. This is the base. Now take 4 triangles such that each of them are connected to one-one side of the base rectangle and when they're taken together, they form a closed object. That shape is called rectangular pyramid.

The base of the pyramid is a rectangle that is not a square.

The pyramid has two planes of symmetry.

An axis of symmetry passes through the vertex and is perpendicular to the base.

An angle that measures 90° is an angle of rotation for an axis of symmetry for the pyramid.

Then the correct options are B, C, and E.

Learn more about rectangular pyramid here:

brainly.com/question/16261758

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7 0

2 years ago

<img src="https://tex.z-dn.net/?f=Given%20%5C%3A%20cotA%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%7D%7B3%7D%7D" id="TexFormula1" title="Giv

ruslelena [56]

<h3>Given :</h3>

$\tt cotA = \sqrt{ \dfrac{1}{3}}$

$\tt \implies cotA = \dfrac{1}{\sqrt{3}}$

All other trigonometric ratios, which are :

sinA
cosA
tanA
cosecA
secA

<h3>Solution :</h3>

Let's make a diagram of right angled triangle ABC.

Now, From point A,

AC = Hypotenuse

BC = Perpendicular

AB = Base

$\tt We \: are \: given, \: cotA = \dfrac{1}{\sqrt{3}}$

$\tt We \: know \: that \: cot \theta = \dfrac{base}{perpendicular}$

$\tt \implies \dfrac{base}{perpendicular} = \dfrac{1}{\sqrt{3}}$

$\tt \implies \dfrac{AB}{BC} = \dfrac{1}{\sqrt{3}}$

$\tt \implies AB = 1x \: ; \: BC = \sqrt{3}x \: (x \: is \: positive)$

Now, by Pythagoras' theorem, we have

AC² = AB² + BC²

$\tt \implies AC^{2} = (1x)^{2} + (\sqrt{3}x)^{2}$

$\tt \implies AC^{2} = 1x^{2} + 3x^{2}$

$\tt \implies AC^{2} = 4x^{2}$

$\tt \implies AC = \sqrt{4x^{2}}$

$\tt \implies AC = 2x$

Now,

$\tt sin \theta = \dfrac{perpendicular}{hypotenuse}$

$\tt \implies sinA = \dfrac{BC}{AC}$

$\tt \implies sinA = \dfrac{\sqrt{3}x}{2x}$

$\tt \implies sinA = \dfrac{\sqrt{3}}{2}$

$\Large \boxed{\tt sinA = \dfrac{\sqrt{3}}{2}}$

$\tt cos \theta = \dfrac{base}{hypotenuse}$

$\tt \implies cosA = \dfrac{AB}{AC}$

$\tt \implies cosA = \dfrac{1x}{2x}$

$\tt \implies cosA = \dfrac{1}{2}$

$\Large \boxed{\tt cosA = \dfrac{1}{2}}$

$\tt tan \theta = \dfrac{perpendicular}{base}$

$\tt \implies tanA = \dfrac{BC}{AB}$

$\tt \implies tanA = \dfrac{\sqrt{3}x}{1x}$

$\tt \implies tanA = \sqrt{3}$

$\Large \boxed{\tt tanA = \sqrt{3}}$

$\tt cosec \theta = \dfrac{hypotenuse}{perpendicular}$

$\tt \implies cosecA = \dfrac{AC}{BC}$

$\tt \implies cosecA = \dfrac{2x}{\sqrt{3}x}$

$\tt \implies cosecA = \dfrac{2}{\sqrt{3}}$

$\Large \boxed{\tt cosecA = \dfrac{2}{\sqrt{3}}}$

$\tt sec \theta = \dfrac{hypotenuse}{base}$

$\tt \implies secA = \dfrac{AC}{AB}$

$\tt \implies secA = \dfrac{2x}{1x}$

$\tt \implies secA = 2$

$\Large \boxed{\tt secA = 2}$

5 0

3 years ago

Read 2 more answers

1. Cindy made a few late payments, and ended up defaulting on her credit card. What would her credit rating be?

Rainbow [258]

Since Cindy’s payments were late and she ended up defaulting her card she would need to improve her credit

4 0

3 years ago

Cameron, Arthur, and Jamie are playing soccer. Their locations are recorded by a motion tracking system. The grid shows distanc

Luden [163]

Answer:

1. Arthur is about 54 meters away from Cameron.

2. Jamie is about 56 meters away from Cameron.

3. Arthur is closer to Cameron.

Step-by-step explanation:

We have been given the locations of Cameron (70,10), Arthur (20,30) and Jamie (45,60) on the the grid in meters.

1. To find the distances of Arthur and Jamie from Cameron we will use distance formula.

$\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

$\text{Distance between Cameron and Arthur}=\sqrt{(70-20)^{2}+(10-30)^{2}}$

$\text{Distance between Cameron and Arthur}=\sqrt{(50)^{2}+(-20)^{2}}$

$\text{Distance between Cameron and Arthur}=\sqrt{2500+400}$

$\text{Distance between Cameron and Arthur}=\sqrt{2900}$

$\text{Distance between Cameron and Arthur}=53.8516480713450403\approx 54$

Therefore, the distance Arthur is about 54 meters away from Cameron.

2. Let us find the distance between Cameron and Jamie.

$\text{Distance between Cameron and Jamie}=\sqrt{(70-45)^{2}+(10-60)^{2}}$

$\text{Distance between Cameron and Jamie}=\sqrt{(25)^{2}+(-50)^{2}}$

$\text{Distance between Cameron and Jamie}=\sqrt{625+2500}$

$\text{Distance between Cameron and Jamie}=\sqrt{3125}$

$\text{Distance between Cameron and Jamie}=55.9016994374947424\approx 56$

Therefore, Jamie is about 56 meters away from Cameron.

3. We can see that 56 is greater than 54, therefore, Arthur is closer to Cameron.

4 0

3 years ago

Read 2 more answers

I also need help with this question.

Softa [21]

1. 4.8 x 10^9
2. 2 x 10^6

5 0

3 years ago

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