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3 years ago

What is answer to 4/5 - 1/10

Mathematics

2 answers:

Zanzabum3 years ago

5 0

Answer: 7/10

0.70000 as a decimal

REY [17]3 years ago

4 0

4/5-1/10=7/10

First, you have to find a common denominator (which is 10 in this case): 5x2=10, and you do the same to the numerator: 4x2=8

This turns the equation into: 8/10-1/10
Since they have the same denominator, you can just subtract the numerator: 8/10-1/10=7/10

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How do you find the distance between two coordinate points on the graph?

borishaifa [10]

Answer:

point of intercept

Step-by-step explanation:

measure x and y axis to see where they cross

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3 years ago

20000 students took a standardized math test. The scores on the test are normally distributed, with a mean score of 85 and a sta

Tpy6a [65]

Answer: 2718

Step-by-step explanation:

Given: Mean score = 85

Standard deviation = 5

Let x be the score of a random student that follows normal distribution.

Then, the probability that a student scored between 90 and 95 will be

$P(90< x < 95)\\\\=P(\dfrac{90-85}{5}$

The number of students scored between 90 and 95 = 0.1359 x (Total students)

= 0.1359 (20000)

= 2718

Hence, The number of students scored between 90 and 95 = 2718

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3 years ago

If ∠1 and ∠2 are complementary and m∠1= 17°, what is m∠2?

hichkok12 [17]

Complementary angles add up to 90*. Therefore, the logical answer would be
D, 77.

5 0

3 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

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

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

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3 years ago

Everett and Marie are going to make brownies for their family reunion. They want to make 4 times the amount of the recipe makes.

KonstantinChe [14]

2/3 x4 = 8/12

therefore,
They will need 8/12 cup oil

3 0

3 years ago