Ask question

Ask question

All categories

2 years ago

12

Tom determines that the system of equations below has two solutions, one of which is located at the vertex of the parabola. Equa

tion 1: (x – 3)2 = y – 4 Equation 2: y = -x + b In order for Tom’s thinking to be correct, which qualifications must be met?

1 answer:

slega [8]2 years ago

7 0

Answer:

b=7

Step-by-step explanation:

You might be interested in

Please help fast<br> The graph shows the function f(x)

DedPeter [7]

Answer:

3/1

Step-by-step explanation:

The average rate of change is the distance between y values divided by the distance between x values. We know this as rise/run or slope for linear equations. For non-linear equations, we calculate the average over an interval.

This function has y-values of -1 at x=0 and 5 at x=2.

We add -1+5=6 and divide this by 2-0=2.

6/2=3.

The y-values on average change 3 units for every 1 unit of the x-values.

4 0

2 years ago

What is the missing number in 12,24,72,288

Ronch [10]

12*2=24
24*3=72
72*4=288
n*(x+1)
288*(4+1)
288*5=1440

5 0

3 years ago

9-3*1/3*98=<br><br> 6-*87/8+85647-552=<br><br><br> 254+968*8

sp2606 [1]

Answer:

don't you have a calculator? just use that

6 0

1 year ago

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

2 years ago

ranslate the sentence into an equation using n as the unknown number. Then solve the equation for n. Two times a number increase

joja [24]

Hi there!

Let's break this down:
Two times a number = 2n
Increased by 5 = + 5
is 15 - = 15

Total equation:
2n + 5 = 15

Solving:
2n + 5 = 15
2n = 10
n = 5

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

4 0

2 years ago