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3 years ago

Creat an expression without parentheses that is equivalent to 4(6x+x)

2 answers:

5 0

You could distribute this expression to get and even MORE simplified expression. Add the two terms in the parentheses together so that you get 4(7x). Now you can simply multiply 4 by 7x which would equal 28x. The new expression is 28x.

Harman [31]3 years ago

3 0

Try to simplify the equation: 4(6x+x). You should distribute first. 24x + 4x. That would equal 28x. If you simplified the equation, you should also get 4(7x). That also equals 28x. 28x is the answer so use 24x + 4x.

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What radian measure is equivalent to 900°? 5n 10n 1/5n 9n

Sergeu [11.5K]

9n is the answer to this question

3 0

3 years ago

Read 2 more answers

!!!!!!!!!!!!help me

steposvetlana [31]

Answer:

Step-by-step explanation:

i'll do a few of them

1) one inch plus a half inch plus an eighth inch plus a sixteenth inch

1 + 1/2 + 1/8 + 1/16 read the tick marks

1 + 8/16 + 2/16 + 1/16 find a common denominators

1 and 11/16 add the sixteenth numerator

4) 6 inch plus an eighth inch plus a sixteenth inch

6 + 1/8 + 1/16 read the tick marks

6 + 2/16 + 1/16 find a common denominators

6 and 3/16 add the sixteenth numerator

5) nine inch plus a quarter inch plus an eighth inch

9 + 1/4 + 1/8 read the tick marks

9 + 2/8 + 1/8 find a common denominators

9 and 3/8 add the sixteenth numerator

5 0

3 years ago

-3 + x = 7.2; x = -4w=-10;w=

inna [77]

Answer:

X=10.2

w= 10/4 or w= 5/2

Step-by-step explanation:

4 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers

A reflecting pool is shaped like a right triangle with one leg along the wall of a building. the hypotenuse is 9 feet longer tha

morpeh [17]

Answer:

Leg side along the wall = x ft = 8 ft

The other leg side = 7+x ft = 7+8=15 ft

The Hypotenuse =9+x ft = 9+8 = 17 ft

Step-by-step explanation:

In the question, the shape of the pool is right triangle.

Let the leg side along the wall to be the x ft

Let the other leg side to be 7+x ft

Let the longest side/hypotenuse to be x+9 ft

Apply the Pythagorean relationship where the sum of squares of the legs equals the square of the hypotenuse

This means;

$x^2 +(x+7)^2=(x+9)^2\\\\$

Expand the terms in brackets

$x^2+(x+7)^2=(x+9)^2\\\\\\x^2+x^2+14x+49=x^2+18x+81$

collect like terms

$x^2+x^2-x^2=18x-14x+81-49\\\\\\x^2=4x+32\\\\\\x^2-4x-32=0$

solve for x in the quadratic equation by factorization

$x^2-4x-32=0\\\\\\x^2-8x+4x-32=0\\\\\\x(x-8)+4(x-8)=0\\\\\\(x+4)(x-8)=0\\\\\\x+4=0,x=-4\\\\x-8=0,x=8$

Taking the positive value of x;

x=8ft

Finding the lengths

Leg side along the wall = x ft = 8 ft

The other leg side = 7+x ft = 7+8=15 ft

The Hypotenuse =9+x ft = 9+8 = 17 ft

6 0

3 years ago