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2 years ago

Someone in Highschool help?

1 answer:

8 0

Solution: y=2x-11

Step by step solution:
f(x)=(x+1)2-7
f(x)=2x+2-7
f(x)=2x-5
y=2x-5
When it’s translated 3 to the right the y intercept is 3 less if the slope is 1, but since the slope is 2 there the difference is 6, so the y intercept is 6 less than it was before
y=2x-11

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5(x + 1) = 4(x + 8)

Reika [66]

<span>5(x + 1) = 4(x + 8)
Use distributive property
5x+5=4x+32
Subtract 4x from both sides
x+5=32
Subtract 5 from both sides
Final Answer: x=27</span>

5 0

2 years ago

Princess Poly claims that the equation (3x−4i)(7x−8)(3x+4i)=0 has three roots, a real root and a complex root with a multiplicit

Serjik [45]

Given equation:

$(3x-4i)(7x-8)(3x+4i)=0$

i.e., $3x-4i=0$ and $7x-8=0$ and $3x+4i=0$

i.e., $3x=4i$ and $7x=8$ and $3x=-4i$

i.e., $x=\frac{4i}{3}$ and $x=\frac{8}{7}$ and $x=-\frac{4i}{3}$

Therefore, the given equation has a real root i.e., $x=\frac{8}{7}$

and two complex conjugate roots i.e., $x=\frac{4i}{3}$ and $x=-\frac{4i}{3}$

Hence, option C is correct.

4 0

2 years ago

Find the probability that a point chosen at random will lie in the shaded region.

malfutka [58]

Step-by-step explanation:

In the 1st question

Total boxes = 6

Shaded boxes = 2

Probability = 2/6 = 1/3

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

2 years ago

Graph matching 50 Points!

Sloan [31]

Step-by-step explanation: To find the graphs, we look at the problem and the graph. Let's do it step-by-step.

No. 1 is y = 1/2.

To find the corresponding graph, we look closely at the graph.

Look at the 2nd graph. Do you see a dot by 1/2? I do!

So, No. 1 matches with the 2nd graph.

No. 2 is y = 2.

Like I said, we look closely at the graph.

Look at the 4th graph. Do you see a dot by 2? I do!

So, No. 2 matches with the 4th graph.

No. 3 is y = 4.

Now, we look closely at the graph.

Look at the 3rd and the 4th graph. Do you see a dot by 4? I do!

So, No. 3 matches with the 3rd and the 4th graph.

Lastly, No. 4 is y = 1/4.

Let's look closely at the graph.

Look at the 1st graph. Do you see a dot by 1/4? I do!

So, No. 4 matches with the 2nd graph.

Therefore, our answers are:

No. 1 matches with the 2nd graph

No. 2 matches with the 4th graph.

No. 3 matches with the 4th and the 3rd graph.

No. 4 matches with the 1st and the 2nd graph.

Download pdf

5 0

2 years ago

Read 2 more answers