Log(x-1)+log(x+2)=log12

ILL GIVE BRAINLEST Determine whether the two angles in each figure are complementary, supplementary, or neither

Firdavs [7]

Answer:

Step-by-step explanation:

If the sum of two angles = 180°

Then the pair of angles will be supplementary.

If the sum of two angles = 90°

Then the pair of angles will be complementary.

Sum of angles having measures 98° and 82° = 98° + 82°

= 180°

Therefore, both the angles will be supplementary.

Sum of angles having measures 134° and 36° = 134° + 36°

= 170°

Therefore, these angles are neither supplementary nor complementary.

So the answer is NEITHER.

8 0

3 years ago

Find the number of faces, edges, and vertices of this shape.

sattari [20]

Faces: 7

Edges: 12

Vertices: 7

3 0

3 years ago

Read 2 more answers

I need help solving number 17.

melisa1 [442]

Let x = amount of sales (in dollars)

The salary is $400 and there's an additional 0.06x dollars added on to get to the goal of 790. The equation is therefore

400+0.06x = 790

Let's solve for x

400+0.06x = 790
400+0.06x-400 = 790-400
0.06x = 390
0.06x/0.06 = 390/0.06
x = 6500

The final answer is 6500

This means he must have $6,500 in sales.

8 0

4 years ago

Y=4x^2-36 solve the following quadratic function by utilizing the square root method x=?

Luda [366]

Answer:

This ans is easily solved by using algebraic identities a²- b² = (a+b) (a-b)

Step-by-step explanation:

1. convert into Identity

2. solve quadratic equation that formed.

7 0

2 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago