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3 years ago

2 and 3/8 minus 1/4 times 1/8

Mathematics

1 answer:

sveta [45]3 years ago

3 0

Answer:

2 11/32 or 75/32

Step-by-step explanation:

2 3/8 - 1/4 x 1/8= 2 3/8 - 1/32= 19/8- 1/32= 76/32 - 1/32= 75/32= 2 11/32

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Suppose that P is the point on the unit circle obtained by rotating the initial ray counterclockwise through 330 degrees. Find t

Flauer [41]

Answer:

Reference angle will be 30°

sin(330°) will be equal to $\frac{-1}{2}$

cos(330°) will be equal to 0.866

Step-by-step explanation:

We have given angle is 330°

As the angle $330^{\circ}$ is in forth quadrant

We know that in forth quadrant the reference angle is given by

Reference angle $=360-angle$

As the angle is 330°

So the reference angle will be $=360-330=30^{\circ}$

Now we have to find the value of

sin(330°) $=sin(2\pi -30)=-sin(30)=\frac{-1}{2}$

And now $cos(330^{\circ})=cos(2\pi -30)=cos30=0.866$

4 0

2 years ago

A spinner marked A, B, C is spun then a 6-sided die is rolled. What is the probability of getting a B and then a 6.

BigorU [14]

Given:

A spinner marked A, B, C is spun then a 6-sided die is rolled.

To find:

The probability of getting a B and then a 6.

Solution:

We know that,

$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

Total possible values for a spinner are 3. So, the probability of getting B, we get

$P(B)=\dfrac{1}{3}$

Total possible values for a die are 6. So, the probability of getting 6, we get

$P(6)=\dfrac{1}{6}$

Now, the probability of getting a B and then a 6 is

$P(B,6)=\dfrac{1}{3}\times \dfrac{1}{6}$

$P(B,6)=\dfrac{1}{18}$

The required probability is $\dfrac{1}{18}$ .

Therefore, the correct option is D.

5 0

2 years ago

Write a definite integral that represents the area of the region. (Do not evaluate the integral.)

joja [24]

Answer:

$\int_{-1}^{1}7(x^{3}-x)\:dx$

Step-by-step explanation:

1) The other curve is $y=0$ then the common points of both curves are x-intercepts, the roots of $y=7(x^{3}-x)$

$y=7(x^{3}-x)\Rightarrow 7(x^3-x)=0 \Rightarrow 7(x^{3}-x)=7x(x-1)(x-1)\Rightarrow \\S=\left ( 0,0 \right ),\left ( -1,0 \right ),\left ( 1,0 \right )$

2). Then those intersection points are the upper and the lower limits. Plugging in to this formula for they belong to the interval [-1,1]:

$\int_{a}^{b}|f(x)-g(x)dx$

$\int_{a}^{b}|f(x)-g(x)dx \Rightarrow \int_{-1}^{1}|7(x^{3}-x)-0|dx \Rightarrow \int_{-1}^{1}7(x^{3}-x)\:dx$

8 0

2 years ago

Find the measures of a positive angle and a negative angle that are coterminal with each given angle

seraphim [82]

Answer:

Why was Mrs Hallette unhappy when people asked about her son?

Step-by-step explanation:

4 0

2 years ago

Multiply.

Crank

The main rules that we use here are :

i) $\sqrt{a \cdot b}= \sqrt{a} \cdot \sqrt{b}$ for nonnegative values a and b.

ii) $\sqrt{a} \cdot \sqrt{a} =a$ .

Thus, first 'decompose' the numbers in the radicals into prime factors:

$4 \sqrt{3} \cdot 10 \cdot \sqrt{2\cdot2\cdot3}\cdot \sqrt{2\cdot 3}\cdot \sqrt{2}.$ .

By rule (i) we write:

$4 \sqrt{3} \cdot 10 \cdot \sqrt{2}\cdot \sqrt{2}\cdot \sqrt{\cdot3}\cdot \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{2}$ .

We can collect these terms as follows:

$40 \sqrt{3}\cdot (\sqrt{3}\cdot \sqrt{3}) \cdot (\sqrt{2}\cdot \sqrt{2}) \cdot( \sqrt{2} \cdot \sqrt{2})$ , and by rule (ii) we have:

$40 \sqrt{3}\cdot 3 \cdot 2 \cdot2=40\cdot12\cdot \sqrt{3}=480 \sqrt{3}.$

Answer: $480 \sqrt{3}$ .

8 0

3 years ago