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Veronika [31]
3 years ago
14

The bar graph shows the number of school days Jalen had homework and did not have homework during the first six months of

Mathematics
1 answer:
Ulleksa [173]3 years ago
8 0

Answer: 3 months

I already did this it’s right

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Thelma Thrifty went to her bank. She had a balance of $1,209.76 in her savings account. She withdrew $131.44 and the teller cred
tangare [24]
The answer is B. $1,084.99
When a bank teller credits money into your account it adds money. And when you withdrawal money it takes it out. So you would set up the problem
$1,209.76-$131.44=$1,078.32
Then add the money credited into the account to that total.
$1,078.32+$6.67=$1,084.99
3 0
4 years ago
Complete the steps in order to write the inverse of f(x) = x + 5
yaroslaw [1]
An inverse function produces points (y,x) for every point (x,y) produced by the parent function.  So to find the inverse function, you solve for the independent variable and when done switch the variable labels....

y=x+5  subtract 5 from both sides

y-5=x  now switch labels

y=x-5

So if f(x)=x+5

f^-1(x)=x-5
7 0
3 years ago
Read 2 more answers
Calculate the volume of the prism below with measurements a = 11, b = 5, and c = 10.
Mariulka [41]
The volume of the prism is 550
4 0
4 years ago
Read 2 more answers
A grocery store is offering a 50% discount off a $4.00 box of cereal. You also have a $1.00 off coupon for the same cereal. Use
s344n2d4d5 [400]
After because if you use it before the discount then the final price will be 1.50 but if you use it after the discount, then the final price is 1 dollar . 4 dollar with 50% discount is 2 dollar, and 2-1 is 1. 4-1 is 3 and 3 dollar with 50% discount is 1.50 dollar
4 0
3 years ago
Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, x = 0^{\circ}

Step-by-step explanation:

Given that : 2\cos^2x -\cos x -1 =0

Let \cos x =y

then our equation become;

2y^2-y-1= 0           .....[1]

A quadratic equation is of the form:

ax^2+bx+c =0.....[2] where a, b and c are coefficient and the solution is given by;

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}

Simplify:

y = \frac{ 1 \pm \sqrt{1+8}}{4}

or

y = \frac{ 1 \pm \sqrt{9}}{4}

y = \frac{ 1 \pm 3}{4}

or

y = \frac{1+3}{4} and y = \frac{1 -3}{4}

Simplify:

y = 1 and y = -\frac{1}{2}

Substitute y = cos x we have;

\cos x = 1

⇒x = 0^{\circ}

and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

The solution set:  \{0^{\circ}, 120^{\circ} , 240^{\circ}\}

Therefore, the solution for the given equation  2\cos^2x -\cos x -1 =0 is, 0^{\circ}





8 0
4 years ago
Read 2 more answers
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