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3 years ago

How many solutions are there to the equation below?

Mathematics

2 answers:

Anton [14]3 years ago

8 0

Answer:

B. Infinitely Many

Step-by-step explanation:

Step 1: Distribute

5x + 48 + 7x = 12x + 48

Step 2: Combine like terms

12x + 148 = 12x + 48

If the 2 lines are the same, which they are, they are the same lines. That means there are an infinite amount of solutions. If they had the same slope but different y-intercepts, there would be no solutions because the lines are parallel. Any other pair of equations would have 1 solution.

Fudgin [204]3 years ago

6 0

Answer:

B. Infinitely Many

Step-by-step explanation:

When you simplify the term on the left of the equation, it is exactly the same as the term on the right. Therefore, you can substitute virtually anything in for variable "x" and it would still be correct.

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Look at the equation below.

USPshnik [31]

Answer:

-10 I think

Step-by-step explanation:

I used a calculator

3 0

2 years ago

10 PIONTS+BRAINLYESS

nikdorinn [45]

Plantation owners wabted runaways slaves

8 0

3 years ago

What is 50 divided by 8675 showing your work

Ierofanga [76]

8675 ÷ .5= 17350
You could also multiply by 2 to check your answer
17350 * .5= 8675
You could also divide by 2.

5 0

2 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

Abby wants to enclose a rectangular area of 20 square feet to use as a garden. Write a function that could be used to find the g

Thepotemich [5.8K]

Xy=20 As the width gets smaller, the length should get larger.

6 0

3 years ago