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1 year ago

Complete the equivalent addition expression for each subtraction

Mathematics

1 answer:

PilotLPTM [1.2K]1 year ago

6 0

Answer:

Step-by-step explanation:

We have to write the additive inverse of the second number.

Additive inverse of 4 is (-4)

Additive inverse of (-4) = 4

A. 23 - 3 = 23 + (-3) = 20

B. 7 - (-15) = 7 + (+15) = 22

C. -15 - 4 = -15 + (-4) = - 19

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A distance of 30 miles on a map is represented by a 2-inch line. If the map is enlarged to 3 times its size, what will be the sc

vichka [17]

Answer:

$\frac{1}{5}\frac{in}{miles}$

Step-by-step explanation:

we know that

The original scale of the map is

$\frac{2}{30}\frac{in}{miles}=\frac{1}{15}\frac{in}{miles}$

If the map is enlarged to 3 times its size

then

the new scale will be

$\frac{2*3}{30}\frac{in}{miles}=\frac{6}{30}=\frac{1}{5}\frac{in}{miles}$

4 0

2 years ago

Suppose that the functions q and r are defined as follows.

satela [25.4K]

Answer:

(r o g)(2) = 4

(q o r)(2) = 14

Step-by-step explanation:

Given

$g(x) = x^2 + 5$

$r(x) = \sqrt{x + 7}$

Solving (a): (r o q)(2)

In function:

(r o g)(x) = r(g(x))

So, first we calculate g(2)

$g(x) = x^2 + 5$

$g(2) = 2^2 + 5$

$g(2) = 4 + 5$

$g(2) = 9$

Next, we calculate r(g(2))

Substitute 9 for g(2)in r(g(2))

r(q(2)) = r(9)

This gives:

$r(x) = \sqrt{x + 7}$

$r(9) = \sqrt{9 +7{$

$r(9) = \sqrt{16}$ {

$r(9) = 4$

Hence:

(r o g)(2) = 4

Solving (b): (q o r)(2)

So, first we calculate r(2)

$r(x) = \sqrt{x + 7}$

$r(2) = \sqrt{2 + 7}$

$r(2) = \sqrt{9}$

$r(2) = 3$

Next, we calculate g(r(2))

Substitute 3 for r(2)in g(r(2))

g(r(2)) = g(3)

$g(x) = x^2 + 5$

$g(3) = 3^2 + 5$

$g(3) = 9 + 5$

$g(3) = 14$

Hence:

(q o r)(2) = 14

8 0

2 years ago

Rewrite in Polar Form.... x^(2)+y^(2)-6y-8=0

skad [1K]

$\bf x^2+y^2-6y-8=0\qquad 
\begin{cases}
x^2+y^2=r^2\\\\
y=rsin(\theta)
\end{cases}\implies r^2-6rsin(\theta)=8
\\\\\\
\textit{now, we do some grouping}\implies [r^2-6rsin(\theta)]=8
\\\\\\\
[r^2-6rsin(\theta)+\boxed{?}^2]=8\impliedby 
\begin{array}{llll}
\textit{so we need a value there to make}\\
\textit{a perfect square trinomial}
\end{array}$

$\bf 6rsin(\theta)\iff 2\cdot r\cdot \boxed{3\cdot sin(\theta)}\impliedby \textit{so there}
\\\\\\
\textit{now, bear in mind we're just borrowing from zero, 0}
\\\\
\textit{so if we add, \underline{whatever}, we also have to subtract \underline{whatever}}$

$\bf [r^2-6rsin(\theta)\underline{+[3sin(\theta)]^2}]\quad \underline{-[3sin(\theta)]^2}=8\\\\
\left. \qquad \right.\uparrow \\
\textit{so-called "completing the square"}
\\\\\\\
[r-3sin(\theta)]^2=8+[3sin(\theta)]^2\implies [r-3sin(\theta)]^2=8+9sin^2(\theta)
\\\\\\
r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}\implies r=\sqrt{8+9sin^2(\theta)}+3sin(\theta)$

3 0

2 years ago

Complete the square. Fill in the number that makes the polynomial a perfect-square quadratic. 2 + 2) +

daser333 [38]

Try using either the app the calculater not like those other ones this calculator can solve fractions and almost any other thing

7 0

3 years ago

Solve the following equation for n. 1) 5(-5 + n) =-3 – 6n A) 2 B) {-9) C) {7} D) {16

emmainna [20.7K]

Answer:

Step-by-step explanation:

5 (-5 + n) = -3 - 6n

-25 + 5n = -3 - 6n

-25 + 11n = -3

11n = 22

n = 2

4 0

2 years ago