All categories

10 months ago

3.5+2=2x-10help me HELP ME JIGGABOOS

Mathematics

1 answer:

Genrish500 [490]10 months ago

5 0

3.5 + 2 = 2x - 10
5.5 = 2x -10
+10 +10
—————————
15.5 = 2x
—— ——
2 2

7.75 = x

7.75 is the answer

You might be interested in

I NEED HELPPPPP!!!! PLZZZ

iVinArrow [24]

Answer:

13.5

Step-by-step explanation:

4 0

2 years ago

Look at the triangle show on the right. The Pythagorean Theorem states that the sum of the squares of the legs of a right triang

Vladimir [108]

Answer:

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

Step-by-step explanation:

Given

$(\frac{b}{r})^2 + (\frac{a}{r})^2$

Required

Use the expression to prove a trigonometry identity

The given expression is not complete until it is written as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = (\frac{r}{r})^2$

Going by the Pythagoras theorem, we can assume the following.

a = Opposite
b = Adjacent
r = Hypothenuse

So, we have:

$Sin\theta = \frac{a}{r}$

$Cos\theta = \frac{b}{r}$

Having said that:

The expression can be further simplified as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$

Substitute values for sin and cos

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$ becomes

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

3 0

3 years ago

Please help me please I only have a couple hours left and this is taking soooo much time

Makovka662 [10]

Answer:

it would be the first answer

Step-by-step explanation:

you can draw the net of the shapes, and count each corner there

3 0

3 years ago

Solve the system of equations. <br> 9x+11y=2 <br> 3x+4y=4

Ronch [10]

Answer:

X= -12 Y=10

Step-by-step explanation:

4 0

2 years ago

I need help writing the first 5 terms of the sequence!

Angelina_Jolie [31]

The first five terms of the sequence are 1, 4, 7, 10, 13.

Solution:

Given data:

$a_{1}=1$

$a_{n}=a_{n-1}+3$

General term of the arithmetic sequence.

$a_{n}=a_{n-1}+d$ , where d is the common difference.

d = 3

$a_{n}=a_{n-1}+3$

Put n = 2 in $a_{n}=a_{n-1}+3$ , we get

$a_{2}=a_1+3$

$a_{2}=1+3$

$a_2=4$

Put n = 3 in $a_{n}=a_{n-1}+3$ , we get

$a_{3}=a_2+3$

$a_{3}=4+3$

$a_3=7$

Put n = 4 in $a_{n}=a_{n-1}+3$ , we get

$a_{4}=a_3+3$

$a_{4}=7+3$

$a_4=10$

Put n = 5 in $a_{n}=a_{n-1}+3$ , we get

$a_{5}=a_4+3$

$a_{5}=10+3$

$a_5=13$

The first five terms of the sequence are 1, 4, 7, 10, 13.

3 0

3 years ago