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Stels [109]
3 years ago
15

I’m confused please help

Mathematics
1 answer:
Maru [420]3 years ago
4 0
<h2>Hello!</h2>

The answer is:

Sin(A)=0.93

<h2>Why?</h2>

To solve the problem, remember to set your calculator in "degree mode"

So, we are given that the angle A is equal to 68°

So, we have that:

Sin(A)=Sin(68\°)=0.93

Hence, we have that:

Sin(A)=0.93

Have a nice day!

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ZP bisects ∠OZQ so that ∠OZP = 8x – 9 and ∠PZQ = 5x + 18. Find the value of x, and the measure of each angle. You must show all
Assoli18 [71]
<span>x = 9 Since ZP bisects â OZQ, that means that the measurements for â OZP and â PZQ are the same. So create an equation with their respective values set to each other. 8x - 9 = 5x + 18 Now solve for x 8x - 9 = 5x + 18 Subtract 5x from both sides 3x - 9 = 18 Add 9 to both sides 3x = 27 Divide both sides by 3 x = 9</span>
6 0
3 years ago
Help me please thank you
oee [108]

Hello from MrBillDoesMath!

Answer:

2(x+8) = 40


Discussion:

Rewrite the original Question as

" Twice the (sum of a number and 8) is 40"   =>

2                ( x                              +    8 ) = 40 =>

2 (x+8) = 40


which is the third bullet point from the top


Thank you,

MrB

7 0
3 years ago
-4 -2 0 2 4 form the set of even integers true or false
GenaCL600 [577]

Answer:

True

Step-by-step explanation:

An integer is anything that's not a fraction/decimal, and being even is common knowledge

3 0
3 years ago
Read 2 more answers
How do i solve for x ?
antoniya [11.8K]

Answer:

Step-by-step explanation:

3 0
3 years ago
A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-t
Vanyuwa [196]

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square  at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

And height will be x

V(x)=x*(26-x)*(13-x)

To get maximum volume differentiate the above equation,

V(x)=x*(26*13-26*x-13*x+x^2)

V(x)=x^3-39x^2+338x\\

V'(x)=3x^2-78x+338

V''(x)=6x-78

Now ,Solve the Quadratic Equation to get x values,

3x^2-78x+338=0

x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

x=[78±Sqrt(3028)]/6

x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17  or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0  i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

V(x)=x^3-39x^2+338x

V(x)=3.8288^3-39*(3.82883)^2+338*3.8288

V(x)=56.13-571.73+1294.13

V(x)=778.53 C.C

6 0
3 years ago
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