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JulsSmile [24]
3 years ago
9

Find the area of the small rectangle. 2x + 3 x + 6 x - 5

Mathematics
1 answer:
Sloan [31]3 years ago
8 0

Answer:

11x-5

Step-by-step explanation:

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Pleaseeeeee help this is due today and I don’t know how to do thissss! Tug of War Activity
Hoochie [10]

Answer:

i

Step-by-step explanation:

7 0
3 years ago
Given: f(n) = 3n-5 and g(n) = 2n-1<br>Find: (fog)(5​
laiz [17]

Answer:

22

Step-by-step explanation:

So we have the two functions:

f(n)=3n-5\text{ and } g(n)=2n-1

And we want to find (fog)(5). This is the same as:

(f\circ g)(5)=f(g(5))

So, find g(5) first:

g(5)=2(5)-1\\

Multiply and subtract:

g(5)=10-1=9

So, we can substitute:

f(g(5))=f(9)

Find the value:

f(9)=3(9)-5

Multiply and subtract:

f(9)=27-5=22

So:

(f\circ g)(5)=22

And we're done!

4 0
3 years ago
Find the area of the right triangle △DEF with the points D (0, 0), E (1, 1), and F.
Tems11 [23]

Answer:

The area of the triangle is  \sqrt{3}

Step-by-step explanation:

Given:

Coordinates D (0, 0), E (1, 1)

Angle  ∠DEF = 60°

△DEF is a Right triangle

To Find:

The area of the triangle

Solution:

The area of the triangle is  = \frac{1}{2}(base \times height)

Here the base is Distance between D and E

calculation the distance using the distance formula, we get

DE  = \sqrt{(0-1)^2 + (0-1)^2}

DE =\sqrt{(-1) ^2 + (-1)^2

DE = \sqrt{1+1}

DE = \sqrt{2}

Base = \sqrt{2}

Height is DF

DF =tan(60^{\circ}) \times DE

DF = \sqrt{3} \times DE

DF = \sqrt{3} \times\sqrt{2}

Now, the area of the triangle is

= \frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})

=\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})

=\frac{1}{2}(2\sqrt{3} )

=\sqrt{3}

6 0
3 years ago
In an 80/20 mortgage, what is the second mortgage used for?
avanturin [10]

Answer:

The first loan covers 80 percent of the home’s price, while the second covers the remaining 20 percent.

Step-by-step explanation:

they are still a home price percentage

5 0
3 years ago
Read 2 more answers
Match the logarithmic functions with their corresponding x-values
cricket20 [7]
We have the following functions:
 log2x=5
 log10x=3
 log4x=2
 log3x=1
 log5x=4
 Let's rewrite each function to solve for x:
 x=2^5=32
 x=10^3=1000
 x=4^2=16
 x=3^1=3
 x=5^4=625
 Answer:
 Matching each function with the solution we have:
 log2x=5 ----------->32
 log10x=3---------->1000
 log4x=2------------>16 
 log5x=4------------>625
3 0
3 years ago
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