MATHEMATICS HIGH SCHOOL

EASY MATH QUESTION PLEASE HELP:
EASY MATH QUESTION PLEASE HELP: - 1

Answers

Answer 1
Answer:

Step-by-step explanation:

a, b, c are real number

Then,

Always True statements are -

  1. abc = cba
  2. a(b + c) = ab + ac
  3. (b - c)a = ab - ac
Answer 2
Answer:

Answer:

ABC are real numbers

Step-by-step explanation:

Related Questions

Factor: 6x+36=  how do I solve this problem
Choose the equivalent system of linear equations that will produce the same solution as the one given below.6x + 2y = −63x − 4y = −1812x + 4y = −1215x = −308x + 4y = −414x = −106x − 8y = −36−6y = −426x − y = −153y = 9
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Mrs. aviles is planning a fruit-cup party for a class of 18 students and two teachers. She spends $10 for a package of snacks and also takes orders in advance for fruit cups that cost $3 each. 1. describe the domain of the function both in words and numbers? 2. what is the range?

Solve for A

W=a/9

A=

Answers

a is w times 9. 

a=W*9.

The value pi/4 is a solution for the equation 3 sqrt 2 cos theta+2=-1

Answers

Answer:

FALSE

Step-by-step explanation:

3\sqrt2\cos\theta+2=-1\n\n\text{Method 1}\n\n\text{Put}\ \theta=(\pi)/(4)\ \text{to the equation and check the equality:}\n\n\cos(\pi)/(4)=(\sqrt2)/(2)\n\nL_s=3\sqrt2\cos(\pi)/(4)+2=3\sqrt2\left((\sqrt2)/(2)\right)+2=((3\sqrt2)(\sqrt2))/(2)+2\n\n=((3)(2))/(2)+2=3+2=5\n\nR_s=-1\n\nL_s\neq R_s\n\n\boxed{FALSE}

\text{Method 2}\n\n\text{Solve the equation:}\n\n3\sqrt2\cos\theta+2=-1\qquad\text{subtract 2 from both sides}\n\n3\sqrt2\cos\theta=-3\qquad\text{divide both sides by}\ 3\sqrt2\n\n\cos\theta=-(3)/(3\sqrt2)\n\n\cos\theta=-(1)/(\sqrt2)\cdot(\sqrt2)/(\sqrt2)\n\n\cos\theta=-(\sqrt2)/(2)\to\theta=(3\pi)/(4)+2k\pi\ \vee\ \theta=-(3\pi)/(4)+2k\pi\ \text{for}\ k\in\mathbb{Z}\n\n\text{It's not equal to}\ (\pi)/(4)\ \text{for any value of }\ k.

Multiply or divide as indicated. Leave your answer with no factors in the denominator.

Answers

Answer:

wheres the question??

Step-by-step explanation:

put a question pls

A regular square pyramid has a base length of 6 cm. The slant height of the pyramid is 9 cm. What is the surface area of the regular pyramid?a. 90 cm2
b. 120 cm2
c. 144 cm2
d. 252 cm2

Answers

l = 6 cm
h = 9 cm
A = l×l + 4×1/2×l×h = 6×6 + 2×6×9 = 36 + 108 = 144 sq cm

HELP MEEEEE PLSSS guysssss

Answers

Answer:

E, 3.

Step-by-step explanation:

Since both f(x) and g[f(x)] are quadratic polynomials, g(x) must also be a linear polynomial.

Let g(x) = Ax + B, where A and B are constants to be determined.

Then we have A[2x² - 3x + 1] + B ≡ x² - (3/2)x + 3.

=> A = 1/2 and B = 5/2.

Hence, f[g(-1)] = f[(1/2)(-1) + (5/2)] = f(2) = 2(2)² - 3(2) + 1 = 3. (E)

Yvette is trying to calculate the distance between point C(1, 2) and point D(7, 10). Which of the following expressions will she use? square root of the quantity of 10 minus 2 all squared plus 7 minus 1 all squared square root of the quantity of 10 minus 1 all squared plus 7 minus 2 all squared square root of the quantity of 10 minus 7 all squared plus 2 minus 1 all squared square root of the quantity of 7 minus 10 all squared plus 1 minus 2 all squared

Answers

So this can be solve using the distance formula: D = sqrt [ (x2 – x1)^2 + (y2 – y1)^2] So the expression she should use is: D = sqrt [ (7 – 1)^2 + (10 – 2)^2] Then the answer would be: D = sqrt [ (6)^2 + (8)^2] D = 10

Answer:

square root of the quantity of 10 minus 2 all squared plus 7 minus 1 all squared

Step-by-step explanation:
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