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2 years ago

Please help with no links.

Mathematics

1 answer:

Wewaii [24]2 years ago

5 0

Answer:

I think its B

I dont know if Im wrong

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Hello, I need help solving the system of equations by substitution method.

mash [69]

4(-5y) + 2y = 180

-20y + 2y = 180

-18y = 180

y = -10
x = 50

5 0

3 years ago

Read 2 more answers

40 POINTS THERE IS MORE THAN 1 ANSWER BRAINLIEST IF CORRECT

mash [69]

Answer:

Whole numbers are also integers. There are other integers which are the opposites of the whole numbers (−1, −2, −3, ...). These negative numbers lie to the left of 0 on the number line. Integers are the whole numbers and their opposites.

4 0

3 years ago

Find the range of values:

Valentin [98]

Answer:

9-10cm

Step-by-step explanation:

You have to find out all the infermation

6 0

4 years ago

James is two years younger than twice his sister Kate’s Age. Together their ages add up to 19. What is each of their ages?

andreev551 [17]

James'age : 12, Kate's age : 7

<h3>Further Explanation </h3>

One variable linear equation is an equation that has a variable and the exponent number is one. Can be stated in the form:

$\large{\boxed{\bold{ax=b}}}$

ax + b = c, where a, b, and c are constants, x is a variable

Whereas the two-variable linear equation is a linear equation that has 2 variables and the exponent is one

Can be stated in the form:

$\large{\boxed{\bold{ax+by=c}}}$

x, y = variable

In the statement of the task:

James is two years younger than twice his sister Kate’s Age can be stated in the equation: James (J) = 2K (Kate) - 2 ⇒equation 1

Together their ages add up to 19. The equation: J + K = 19 .... 2 ---> J = 19 - K ⇒equation 2

So the solution is:

Equation 2 is entered into equation 1 so that it becomes:

19-K = 2K-2

21 = 3K

K = 7

because J + K = 19, then the value of J = 19-7 = 12

<h3>Learn more</h3>

the slope-intercept form of the linear equation

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system of linear equations

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Keywords : Age, variable, linear equation

4 0

4 years ago

If you are reading this please help!!!!

Soloha48 [4]

Given:

$1. -3 x^{2}\left(4 x^{3}-7\right)\\$2. $(6 x-5)(2 x+3)$\\3. $(3 x-1)\left(x^{2}+5 x-2\right)$$

To find:

The product of the polynomials.

Solution:

1. $-3 x^{2}\left(4 x^{3}-7\right)$ $=-3 x^{2}(4 x^{3}) -3 x^{2}(-7)$

Multiply the numerical coefficient and add the powers of x.

$=-12 x^{5}+21 x^{2}$

2. $(6 x-5)(2 x+3)=6 x(2 x+3)-5(2x+3)$

Multiply each term of first polynomial with each term of 2nd polynomial.

Multiply the numerical coefficient and add the powers of x.

$=12 x^2+18x-10 x-15$

$=12 x^2+8x-15$

3. $(3 x-1)\left(x^{2}+5 x-2\right)=3 x(x^{2}+5 x-2)-1(x^{2}+5 x-2)$

Multiply each term of first polynomial with each term of 2nd polynomial.

Multiply the numerical coefficient and add the powers of x.

$=3x^{3}+15 x^2-6x-x^{2}-5 x+2$

Add or subtract like terms together.

$=3x^{3}+14 x^2-11x+2$

The answer for multiplying polynomials:

$1. -12 x^{5}+21 x^{2}$

$2. \ 12 x^2+8x-15$

$3. \ 3x^{3}+14 x^2-11x+2$

3 0

3 years ago