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3 years ago

If h(x)=5x-3 find h(-4)

Mathematics

2 answers:

Iteru [2.4K]3 years ago

5 0

5(-4)-3
-20-3
-23
x=-23

Papessa [141]3 years ago

4 0

In order to gind h (-4), you simply substitute -4 in place of any x in the function.
h(-4) = 5 (-4) - 3
= -20 - 3
= -23

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What is the distance between (9,-4) and (-22, 8)?

MariettaO [177]

Answer:

≈ 33.2 units

Step-by-step explanation:

Calculate the distance d using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = (9, - 4) and (x₂, y₂ ) = (- 22, 8)

d = $\sqrt{(-22-9)^2+(8+4)^2}$

= $\sqrt{(-31)^2+12^2}$

= $\sqrt{961+144}$

= $\sqrt{1105}$ ≈ 33.2 ( to 1 dec. place )

3 0

3 years ago

F=(9/5) C +32, C=15 Can you show me what a graph looks like?

CaHeK987 [17]

Answer:

Step-by-step explanation:

F=(9/5) C +32 has the form of the equation of a straight line: y = mx + b, where m is the slope and b is the y-intercept. By comparing these two equations we see that the slope, m, is (9/5) and the y-intercept is 32 (or (0, 32)).

To graph this, first locate the y-intercept (0, 32); put a dark dot there. Since the slope is 32, or 32/1, move your pencil point 1 unit to the right, from (0, 32) to (1, 32), and then from (1, 32) move y our pencil point 32 units upward. Plot another dark dot there and then draw a straight line through these two points.

We can evaluate the given function F=(9/5) C +32 at C = 15 by replacing C in the formula with 15: F = (9/5)(15) + 32, or F = 59.

8 0

3 years ago

Use compatible numbers to estimate the quotient 315 divided by 8 answer quickly thank you!

lbvjy [14]

315 is about 320 which is divisible by 8 320/8 is 40

8 0

3 years ago

Read 2 more answers

How many Solutions does this system have? (1 point)

mixas84 [53]

The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$