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3 years ago

15

I need help with my IXL long terms. Can someone help/tell me how to do them????

1 answer:

ivanzaharov [21]3 years ago

8 0

The answer to this is y=x+5

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1. Evaluate the expression for the given value of the variable. 4(z+ 4) + 5 for z = 2

Lunna [17]

Answer:

D

Step-by-step explanation:

4(z+ 4) + 5 z = 2

4(2+4)+5

4(6)+5

24+5

=29

7 0

3 years ago

Read 2 more answers

Leani makes a slice that results in a square. What slice did Leani make?

Thepotemich [5.8K]

A cut that is parallel to the base of the square based pyramid

7 0

4 years ago

The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0

3 years ago

S3 > 125<br><br> Solve this problem

Iteru [2.4K]

Use the commutative property to record the terms. Divide both sides inequality by 3 then it’s 3 > 125/3

3 0

4 years ago

Read 2 more answers

Use the parabola tool to graph the quadratic function f(x)=x2−12x+27. Graph the parabola by first plotting its vertex and then p

alina1380 [7]

Answer:

vertex is (6 , -9)

points are (9,0) and (3,0)

Step-by-step explanation:

Given quadratic function $f(x)=x^2-12x+27$

We have to plot the given quadratic function.

Consider the Given quadratic function $f(x)=x^2-12x+27$

The general form of quadratic function is given $f(x)=a(x-h)^2+k$

Where, (h, k) is vertex , given $h =\frac{-b}{2a}$ and $k = f(h)$

Thus, for given quadratic function $f(x)=x^2-12x+27$

a = 1 , b= -12 , c = 27

Thus,

$h =\frac{-b}{2a}=\frac{12}{2}=6$

$k = f(h)$ that is f(12) = (6)^2 - 12× 6 +27 = 36 - 72 + 27 = - 9

Thus, given quadratic function $f(x)=x^2-12x+27$ in standard form is $f(x)=(x-6)^2-9$

Thus, vertex is (6 , -9)

For second point put $f(x)=0$ , we get,

$f(x)=(x-6)^2-9=0$

$\Rightarrow (x-6)^2-9=0$

$\Rightarrow (x-6)^2=9$

$\Rightarrow (x-6)=\pm 3$

$\Rightarrow x= 6\pm 3$

Thus, $\Rightarrow x=6+3=9$ and $\Rightarrow x=6-3=3$

thus, points are (9,0) and (3,0)

Graph is attached below.

5 0

4 years ago