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patriot [66]
3 years ago
12

B is the factor of ab+bc. What is the other factor?

Mathematics
1 answer:
ryzh [129]3 years ago
7 0

Answer:

cannot be simplified anymore

Step-by-step explanation:

b(a+c)

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melamori03 [73]
Just plug in a calculator 336/7 and you should get a non decimal answer.
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3. Fill in the blanks with center, circle, directrix, focus, line, parabola, or radius.
anzhelika [568]

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3 years ago
Read 2 more answers
1st. (2 root 2 + root 3 ) ( 2 root 3 - root 2) 2nd.(root 5 + 2 root 10) (3 root 5 + root 10) 3rd.(4 root 6 - 3 root 3) (2 root 3
Viefleur [7K]

Answer:

1. 3√6 + 10

2. 35 + 7√50

3. 24√2 - 20√6 + 15√3 - 18

4. 17√6 - 38

5. 13√10 - 42

Step-by-step explanation:

The questions are in surd form.

1. (2√2 + √3)(2√3 - √2)

open the parenthesis by multiplying both together .

4√6 + 4 + 6 - √6

3√6 + 10

2. (√5 + 2√10)(3√5 + √10)

15 + √50 + 6√50 + 20

35 + 7√50

3. (4√6 - 3√3)(2√3 - 5)

8√18 - 20√6 - 18 + 15√3

8√9 × 2 - 20√6 - 18 + 15√3

24√2 - 20√6 + 15√3 - 18

4. (6√3 - 5√2)(2√2 - √3)

12√6 - 18 - 20 + 5√6

17√6 - 38

5 (√10 - 3)(4 - 3√10)

4√10 - 30 - 12 + 9√10

13√10 - 42

5 0
3 years ago
Find the circumference and area of a circle with the diameter AB,<br> A(-8,4) and B(4,-1)
bezimeni [28]

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-8}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{4}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AB=\sqrt{[4-(-8)]^2+[-1-4]^2}\implies AB=\sqrt{(4+8)^2+(-1-4)^2} \\\\\\ AB=\sqrt{144+25}\implies AB=\sqrt{169}\implies \stackrel{\textit{diameter}}{AB=13}~\hfill \boxed{\stackrel{\textit{radius}}{6.5}} \\\\[-0.35em] ~\dotfill

\bf \textit{circumference of a circle}\\\\ C=2\pi r\qquad \qquad C=2\pi (6.5)\implies C=13\pi \implies C\approx 40.84 \\\\\\ \textit{area of a circle}\\\\ A=\pi r^2\qquad \qquad A=\pi (6.5)^2\implies A=42.25\pi \implies A\approx 132.73

5 0
3 years ago
Someone help please thank you
mojhsa [17]

Answer:

See below.

Step-by-step explanation:

So we have the two functions:

f(x)=8x-5\text{ and } g(x)=9-2x

And we want to find:

(f\circ g)(x)\text{ and } (g\circ f)(x)

1)

Recall that:

(f\circ g)(x)

is the same as:

=f(g(x))

Thus, we can substitute g(x):

=f(9-2x)

And substitute that into f(x):

f(x)=8x-5\\f(9-2x)=8(9-2x)-5

Distribute:

=72-16x-5

Subtract and simplify:

=67-16x\\=-16x+67

Thus:

(f\circ g)(x)=-16x+67

2)

Similarly:

(g\circ f)(x)=g(f(x))

Substitute f(x):

g(f(x))=g(8x-5)

Substitute:

g(8x-5)=9-2(8x-5)

Distribute:

=9-16x+10

Simplify:

=-16x+19

Therefore:

(g\circ f)(x)=-16x+19

4 0
4 years ago
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