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2 years ago

Evaluate each expression when x = 8, and z = 10

Mathematics

1 answer:

sveticcg [70]2 years ago

7 0

Answer:

x = 2

y = 4

z = 8

1. xy

2. x + z

3. z - y

4. x + y + z

14Step-by-step explanation:

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Write down a pair of integers whose

liubo4ka [24]

The pair of integers that I chose are:

(a) sum is –3
5 + (-8) = -3

(b) difference is –5
2 - 7 = -5

(c) difference is 2
14 -12 = 2

(d) sum is 0
2 - 2 = 0

3 0

3 years ago

Norm's electric bill for December 2013 was $149.32. In 2014 the electric bill was $194.31. Next year Norm expects the bill to be

docker41 [41]

A): 2013 = 149.32

Find difference of: 194.31

2014 = 194.31 149.32

= 44.99

2015 Total Will Be:

B):

149.32 + 25 = 174.32

(2013)

(B) (A)

6 0

1 year ago

Which is the graph of the following equation?

lesantik [10]

Given equation is

$\frac{(x-5)^2}{16} - \frac{(y-2)^2}{9}=1$

The given equation is in the form of

$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1$

a^2= 16 , so a=4

b^2 = 9 so b= 4

The value of 'a' is greater than the value of 'b'

So it is a Horizontal hyperbola

First two graphs are horizontal hyperbola

Here center is (h,k)

h= 5 and k =2 from the given equation

So center is (5,2)

Now we find vertices

Vertices are (h+a,k) and (h-a,k)

We know h=5, k=2 and a=4

So vertices are (9,2) and (1,2)

Second graph having same vertices and center

The correct graph is attached below

7 0

3 years ago

Read 2 more answers

Bob says that when he multiplies (x+3) (x-3) he gets x^2-6x-9 is he correct explain

almond37 [142]

(x+3)(x-3)
x*x=x^2
x*-3=-3x
3*x=3x
3*-3=-9

where’s where he most likely messed up
x^2-3x+3x-9

the correct answer is x^-9

5 0

3 years ago

Create the equation of a quadratic function with a vertex of (5,6) and a y-intercept of -69

Elena L [17]

If the y-intercept is at -69, meaning the point is (0, -69), thus x = 0, y = -69

$\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
vertex~(5,6)\quad 
\begin{cases}
x=5\\
y=6
\end{cases}\implies y=a(x-5)^2+6
\\\\\\
\textit{we also know that }
\begin{cases}
x=0\\
y=-69
\end{cases}\implies -69=a(0-5)^2+6
\\\\\\
-75=25a\implies \cfrac{-75}{25}=a\implies a=-3
\\\\\\
therefore\qquad \boxed{y=-3(x-5)^2+6}$

3 0

2 years ago