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2 years ago

8

If in the equation [ 4/t-1 = 2/w-1 ] t ≠ 1 and w ≠ 1, then t =

1 answer:

schepotkina [342]2 years ago

3 0

t=2w-1

OPTION A is the correct answer...

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The diameter of a tire is 2.5 ft. Use this measurement to answer parts a and b. Show all work to receive full credit.

konstantin123 [22]

Thank you for posting your question here. Please see below answers for the problem above:

a. to find the circumference of the tire you need to multiply 3.14 by 2.5:
= 3.14 x 2.5
= 7.85 to round of it will be 8

b. to get the number of times the tire have to rotate to travel 1 mile:

1 mile = 5,280 ft
1 rotation is 7.85
5,280/7.85 = 672.611465

4 0

2 years ago

Jenny bakes 10 cookies. She puts 7 chocolate chips on each cookie. Draw a tape diagram, and label the total amount of chocolate

IgorC [24]

Jenny bakes 10 cookies. she puts 7 chocolate chips on each cookie draw a tape diagram and label the total amount

3 0

2 years ago

Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

4 0

3 years ago

And Again|||||||||||||||||||||||||||||||||

bogdanovich [222]

Answer:

free points TYSM!

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

In equilateral ΔABC, AD, BE, and CF are medians. If FO = 4, then AE = A) 4 B) 4 3 C) 8 D) 12

vampirchik [111]

Firstly, we need to draw triangle

we know that

O is a centroid

and centroid divides median into 2:1

so,

$\frac{CO}{FO} =\frac{2}{1}$

we have FO=4

so, we can plug it

$\frac{CO}{4} =\frac{2}{1}$

$CO=8$

now, we can find CF

CF=OC+FO

CF=8+4

CF=12

now, we can see triangle ACF is a right angled triangle

so, we can use pythagoras theorem

$x^2=(\frac{x}{2} )^2+12^2$

now, we can solve for x

$\frac{3x^2}{4}=144$

$x^2=192$

$x=8\sqrt{3}$

Since, it is equilateral triangle

so,

$AC=x=8\sqrt{3}$

we know that

E is a mid-point

so,

$AE=\frac{AC}{2}$

now, we can plug values

$AE=\frac{8\sqrt{3}}{2}$

$AE=4\sqrt{3}$ ................Answer

5 0

2 years ago

Read 2 more answers