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3 years ago

What expression is equivalent to -1/3(6x+15)-3

Mathematics

1 answer:

Anna11 [10]3 years ago

5 0

Answer:

$\large\boxed{-\dfrac{1}{3}(6x+15)-3=-2x-8}$

Step-by-step explanation:

$-\dfrac{1}{3}(6x+15)-3\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\=\left(-\dfrac{1}{3}\right)(6x)+\left(-\dfrac{1}{3}\right)(15)-3\\\\=-2x-5-3\\\\=-2x-8$

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Mr. Hammond earns $13.75 per hours and worked 20.5 hours last week. How much did he earn last week?

IRISSAK [1]

Multiply total hours worked by pay rate:

20.5 x $13.75 = $281.88

8 0

3 years ago

I am donating packages for Christmas drive. After the first week, I spent 126 to donate three children packages and 8 adult pack

anastassius [24]

Answer:

Step-by-step explanation:

Let the price for children and adults package be represented with x and y respectively.

For the first week the sum of the package will be as follows:

3x+8y = 126

For the two weeks after:

6x+4y= 108

So, we will be having two equations

3x+8y = 126..... (1)

6x+4y= 108.......(2)

These are simultaneous equations

From equation 1

3x+8y = 126

3x = 126-8y

X = 126-8y/3 ............. (3)

Put equation 3 into 2

6 ( 126-8y)/3 +4y = 108

756-48y/3 +4y = 108

756-48y+12y/3 = 108

Cross multiplying

756-48y+12y= 108×3

756-48y+ 12y = 324

Collecting like terms

756-324 = 48y-12y

432= 36y

Divide both sides by 36

432/36 = 36y/36

y= 12

Substituting y into equation 1

3x+8= 126

3x+96=126

3x= 126-96

3x= 30

Divide both sides by 3

3x/3 = 30/3

x = 10

Hence for each of the packages for children and adults. It will be 10 and 12 respectively.

8 0

3 years ago

8 divided by (7 - 3) x ( 4 + 6)<br> Equals?

gregori [183]

8÷(7-3)x(4+6)

8÷4x10

2x10

3 0

3 years ago

(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0

Goshia [24]

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

$P(A)+P(B)+P(A\cap B)=P(A\cup B)$

We know that

Maximum value of $P(A\cup B)$ =1 and minimum value of $P(A\cup B)$ =0

$0\leq P(A\cup B )\leq 1$

$0\leq P(A)+P(B)-P(A\cap B)\leq 1$

$0\leq 0.4+0.7-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)$

$P(A\cap B)\leq 1.1$

It is not possible that $P(A\cap B)$ is equal to 1.1

$1.1-P(A\cap B)\leq 1$

$-P(A\cap B)\leq 1-1.1=-0.1$

Multiply by (-1) on both sides

$P(A\cap B)\geq 0.1$

Again, $P(A\cup B)\geq P(B)$

$0.4+0.7-P(A\cap B)\geq 0.7$

$1.1-P(A\cap B)\geq 0.7$

$-P(A\cap B)\geq -1.1+0.7=-0.4$

Multiply by (-1) on both sides

$P(A\cap B)\leq 0.4$

Hence, $0.1\leq P(A\cap B)\leq 0.4$

3 0

3 years ago

The quotient of 5 plus d and 12 minus w

aev [14]

That expression is written like so:

(5 + d)/(12 - w)

6 0

3 years ago