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Harman [31]
3 years ago
10

Find the missing dimensions of each triangle described. height 14 in. area: 245 in2

Mathematics
1 answer:
Gemiola [76]3 years ago
7 0
A= \frac{1}{2}bh
b= \frac{2A}{h}= \frac{2(245)}{14}=35

The base of the triangle is 35 in.
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3 years ago
(cot^2x - 1)/(csc^2x) = cos2x​
Alex787 [66]

Answer:

Step-by-step explanation:

The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):

cos(2x)=cos^2x-sin^2x,

cos(2x)=1-2sin^2x, and

cos(2x)=2cos^2x-1

We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants.  Rewriting gives you:

\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }

Notice I also wrote the 1 in terms of sin^2(x).

Now we will put the numerator of the bigger fraction over the common denominator:

\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }

The rule is bring up the lower fraction and flip it to multiply, so that will give us:

\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}

And canceling out the sin^2 x leaves us with just

cos^2x-sin^2x which is one of our identities.

5 0
3 years ago
To celebrate the first day of school, Rita brought in a tray of caramel brownies and walnut brownies. She brought 16 caramel bro
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Answer:

40% were caramel

Step-by-step explanation:

(pls mark brainliest)

5 0
3 years ago
Use substitution to solve for x in the system of equations.
jolli1 [7]
Equation 1) 10x + 2y = 30
Equation 2) 4x + y = 4

Multiply all of equation 2 by 2.

2)  2(4x + y = 4)

2)  8x + 2y = 8
1)  10x + 2y = 30

Subtract equations from each other.

-2x = -22

Divide both sides by -2.

x = -22/-2

x = 11

B) x = 11

~Hope I helped!~


6 0
3 years ago
(2x^2+4x-3)-(2x^2+4x-3) show work mark brainliest
Aleksandr [31]

Solution: \left(2x^2+4x-3\right)-\left(2x^2+4x-3\right)=0

Steps:

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a, =2x^2+4x-3-\left(2x^2+4x-3\right)

-\left(2x^2+4x-3\right):\quad -2x^2-4x+3

=2x^2+4x-3-2x^2-4x+3

\mathrm{Simplify}\:2x^2+4x-3-2x^2-4x+3:\quad 0

=0

8 0
3 years ago
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