All categories

3 years ago

Any help? Please!!! It's due today.

Mathematics

1 answer:

tangare [24]3 years ago

5 0

Answer:

120

Step-by-step explanation:

4+4=8 tell you het to 120

You might be interested in

Evaluate 2b² - 3b + 4a² for a= -6 and b=8

eduard

Answer:

Step-by-step explanation:

2b^2 - 3b + 4a^2........a =-6 and b = 8

now we sub

2(8^2) - 3(8) + 4(-6^2)

2(64) - 24 + 4(36)

128 - 24 + 144

272 - 24

248 <==

7 0

3 years ago

Lionel wanted to share 4 muffins with his 5 friends. He separated the muffins into 6 equal parts

Alinara [238K]

Answer:

24 muffins, each friend gets 4. There are 4 remaining

Step-by-step explanation:

4 0

3 years ago

Invested $8200 at the rate of 4.5% p.a. It earned $738 simple interest. The

Igoryamba

Answer:

2 yrs

Step-by-step explanation:

SI =PTR/100

738=8200 x T x 4.5 /100

738 x 100 / 8200 x 4.5 = T

T =2

3 0

3 years ago

What is the solution to the system of equations below?

katen-ka-za [31]

$4x - y = - 5 \\ - 2x + y = 3$

Solve the second equation for y

$4x - y = - 5 \\ y = 3 + 2x$

Substitute the given value of y into the first equation

$4x - (3 + 2x) = - 5$

Solve the equation for x

$x = - 1$

Substitute the given value of x into the second equation

$y = 3 + 2( - 1)$

Solve the equation for y

$y = 1$

The possible solution of the system is the ordered pair (x,y)

$(x,y) = ( - 1, 1)$

7 0

3 years ago

Find all solutions of the equation in the interval [0, 2pi).

natali 33 [55]

Answer:

Step-by-step explanation:

Begin by squaring both sides to get rid of the radical. Doing that gives you:

$sin^2x=1-cosx$

Now use the Pythagorean identity that says

$sin^2x =1-cos^2x$ and make the replacement:

$1-cos^2x=1-cosx$ . Now move everything over to one side of the equals sign and set it equal to 0 so you can factor:

$1-cos^2x+cosx-1=0$ and then simplify to

$cosx-cos^2x=0$

Factor out the common cos(x) to get

$cosx(1-cosx)=0$ and there you have your 2 trig equations:

cos(x) = 0 and 1 - cos(x) = 0

The first one is easy enough to solve. Look on the unit circle and see where, one time around, where the cos of an angle is equal to 0. That occurs at

$x=\frac{\pi }{2},\frac{3\pi}{2}$

The second equation simplifies to

cos(x) = 1

Again, look to the unit circle and find where the cos of an angle is equal to 1. That occurs at π only.

So, in the end, your 3 solutions are

$x=\frac{\pi}{2},\pi,\frac{3\pi}{2}$

8 0

3 years ago