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3 years ago

Show that sec^2x+csc^2x/sec^2xcsc^2x

Mathematics

1 answer:

professor190 [17]3 years ago

4 0

Answer:

$\frac{1}{csc^{2}(x)}+\frac{1}{sec^{2}(x)}$

Step-by-step explanation:

The image below shows step-by-step on how to solve it.

Hope this helps! :)

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21 and 22 are complementary angles. The measure of Z1 is 45 The measure of Z2 is 3x°. Find the value of x. The figure is not dra

Alex

9514 1404 393

Answer:

x = 15

Step-by-step explanation:

Given:

∠1 + ∠2 = 90°

∠1 = 45°

∠2 = 3x°

Find:

Solution:

Using the given values in the given equation, we have ...

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2 years ago

You plant a rectangular rose garden along the side of your garage. You enclose 3 sides of the garden with 40 feet of fencing. Th

solmaris [256]

Let's assume

length of rectangle =L

width of rectangle =W

You enclose 3 sides of the garden with 40 feet of fencing

so, we get

$L+2W=40$

now, we can solve for L

$L=40-2W$

we know that

area of rectangle is

$A=L*W$

$100=L*W$

now, we can plug

$100=(40-2W)*W$

now, we can solve for W

$-2W^2+40W-100=0$

we can use quadratic formula

$W=\frac{-40+\sqrt{40^2-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}$

$W=\frac{-40-\sqrt{40^2-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}$

we can take anyone value ..because both are giving positive value

first dimensions:

$W=2.929$

now, we can find L

$L=40-2*2.929$

$L=34.142$

so, length is 34.142feet

width is 2.929 feet

Second dimensions:

$W=17.071$

now, we can find L

$L=40-2*17.071$

$L=5.858$

so, length is 5.858feet

width is 17.071 feet

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3 years ago