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3 years ago

Solve using substitution

Mathematics

1 answer:

I am Lyosha [343]3 years ago

6 0

Hopefully, you can understand this!

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Which equation describes the circle having center point (3,7) and radius r = 4

Mnenie [13.5K]

Step-by-step explanation below

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2 years ago

Factorize 2x² - 1 = 0<br><br> Thanks!

erma4kov [3.2K]

2 x2-1,. 2x2=1 or all in the photo any problem in question then ask me

7 0

3 years ago

(100 POINTS)

Sav [38]

Answer:

1. $\frac{7^8}{7^6}=49$

2. 7

3. 18

Step-by-step explanation:

Question 1:

Given:

The expression to simplify is $\frac{7^8}{7^6}$ .

We use the property of exponent $\frac{a^m}{a^n}=a^{m-n}$

Therefore, $\frac{7^8}{7^6}=7^{8-6}=7^2=7\times 7 = 49$

Question 2:

Given:

The expression to simplify is $7(\frac{1}{2}+\frac{1}{2})$ .

We add the fractions first inside the parenthesis.

So, $\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1$

Now, $7(1)=7$

Question 3:

The first five prime numbers are 1, 2, 3, 5, and 7.

Therefore, the sum of these five numbers is given as:

$1+2+3+5+7=18$

5 0

4 years ago

Read 2 more answers

Find the exact value of the trigonometric expression given that sin(u) = -(3/5),

GuDViN [60]

Recall that

$\cot(u+v)=\dfrac{\cos(u+v)}{\sin(u+v)}=\dfrac{\cos u\cos v-\sin u\sin v}{\sin u\cos v+\cos u\sin v}=\dfrac{\cot u\cot v-1}{\cot v+\cot u}$

Also, recall that for all $\theta$ ,

$\cos^2\theta+\sin^2\theta=1$

With $\frac{3\pi}2$ , we can expect $\cos u>0$ , and with $0$ , $\sin v>0$ . So from the above identity it follows that

$\cos u=\sqrt{1-\sin^2u}=\dfrac45\implies\cot u=\dfrac{\cos u}{\sin u}=-\dfrac43$

and

$\sin v=\sqrt{1-\cos^2v}=\dfrac8{17}\implies\cot v=\dfrac{\cos v}{\sin v}=\dfrac{15}8$

and so

$\cot(u+v)=\dfrac{-\frac43\frac{15}8-1}{\frac{15}8-\frac43}=\boxed{-\dfrac{84}{13}}$

6 0

3 years ago

Please help!!! 25 points!!

Eduardwww [97]

<h3>Answer: B) y = -x</h3>

==============================================

Work Shown:

g(x) = x^2-2

g(-2) = (-2)^2-2

g(-2) = 2

Saying (-2, g(-2)) is the same as saying (-2, 2)

---------------

g(x) = x^2-2

g(1) = 1^2 - 2

g(1) = -1

Saying (1, g(1)) is the same as saying (1, -1)

---------------

Let's find the slope of the line through (-2,2) and (1,-1)

m = (y2 - y1)/(x2 - x1)

m = (-1-2)/(1-(-2))

m = (-1-2)/(1+2)

m = -3/3

m = -1

Now turn to point slope form to find the equation

y - y1 = m(x - x1)

y - 2 = -1(x - (-2))

y - 2 = -(x + 2)

y - 2 = -x - 2

y-2+2 = -x-2+2 ... add 2 to both sides

y = -x

You could use the other point (1,-1) and you'd get the same answer. The slope m = -1 is the same each time.

7 0

4 years ago