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igor_vitrenko [27]

3 years ago

8

Find the product of 3(x+4)(x-5)

2 answers:

valentinak56 [21]3 years ago

8 0

First distribute the 3
(3x + 12)(x - 5)
Multiply
3x ${^2}$ - 3x - 60

ASHA 777 [7]3 years ago

4 0

For this case we have the following steps:

Step 1:

We apply distributive property, which states:

$a (b + c) = ab + ac$

Applying the property to the product of "3" and the first parenthesis we have:

$(3x + 12) (x-5)$

Step 2:

Apply distributive property again

$3x ^ 2-15x + 12x-60$

Step 3:

We add similar terms

$3x ^ 2-3x-60$

Thus, the product is given by: $3x ^ 2-3x-60$

Answer:

$3x ^ 2-3x-60$

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Julli [10]

Answer:

g(2) = 5

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g(5) = 1

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Step-by-step explanation:

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4 0

3 years ago

What points are on the graph -3x+6y+5=-7

Simora [160]

The slope 1/2 y -intercept -2

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3 years ago

If nelson invests $5000 for 4 years and earns $600, what is the simple interest rate?

klio [65]

Answer: Simple interest rate = 3%.

Step-by-step explanation: We are given invested amount = $5000.

Number of years of investment = 4 years.

Interest earned = $600.

Simple interest formula.

I = P R T.

Where I is the earned interest, P is the invested amount , R is the rate of interest and T is the time in years.

Plugging values in formula, we get

600 = 5000 × R × 4

600 = 20000R.

Dividing both sides by 20000, we get

$\frac{600}{20000} = \frac{20000R}{20000} .$

0.03 = R.

Multiplying by 100 to convert it in percentage, we get

0.03 × 100 = R

R = 3 %.

4 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

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marusya05 [52]

the answer is 48-8f

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3 years ago