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3 years ago

Math again help please!!

Mathematics

1 answer:

katrin [286]3 years ago

3 0

I would say they would need to be replaced at separate times. If the tires diameter is greater then the other can travel double the distance without needing to be changed. Thats what i think

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Why is the answer 2 in the subtractive equation 1+1?

Yuki888 [10]

Answer:

That's not a subtractive equation, the symbol that you used indicates that it is an addition equation. if the question was 1-1 then the answer wouldn't be 2 but since the question is 1+1 the answer is 2

Step-by-step explanation:

6 0

2 years ago

Compare partial products and regrouping

Stella [2.4K]

Let us first describe how partial product and regrouping are alike:

Partial Products and Regrouping are alike because both methods are multiplied by one number and if the product of the number has 2 digits it can be carried.

Now let us discuss how they are different:

Partial Products and Regrouping are different because Partial Products are doing multiplication step by step and regrouping is regular multiplication.

Hope it helps.

5 0

3 years ago

Wire A is bent to form a rectangle with dimensions 5x cm by (4x +2) cm. Wire B is bent to form another rectangle with dimensions

Masja [62]

Answer:

Rectangle B has greater perimeter.

3 0

2 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

Desiree created a website with separate pages for videos, games, photos and poems. She made the table below to

dusya [7]

Answer: 3 to 2

Step-by-step explanation:

You simplify the ratio :)

3 0

2 years ago