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Roman55 [17]
2 years ago
10

Pls answer a,b,2,3,4

Mathematics
1 answer:
MArishka [77]2 years ago
3 0

The writing isnt so clear to me
You might be interested in
1. Consider the right triangle ABC given below.
lbvjy [14]
#1) 
A) b = 10.57
B) a = 22.66; the different methods are shown below.
#2)
A) Let a = the side opposite the 15° angle; a = 1.35.
Let B = the angle opposite the side marked 4; m∠B = 50.07°.
Let C = the angle opposite the side marked 3; m∠C = 114.93°.
B) b = 10.77
m∠A = 83°
a = 15.11

Explanation
#1)
A) We know that the sine ratio is opposite/hypotenuse.  The side opposite the 25° angle is b, and the hypotenuse is 25:
sin 25 = b/25

Multiply both sides by 25:
25*sin 25 = (b/25)*25
25*sin 25 = b
10.57 = b

B) The first way we can find a is using the Pythagorean theorem.  In Part A above, we found the length of b, the other leg of the triangle, and we know the measure of the hypotenuse:
a²+(10.57)² = 25²
a²+111.7249 = 625

Subtract 111.7249 from both sides:
a²+111.7249 - 111.7249 = 625 - 111.7249
a² = 513.2751

Take the square root of both sides:
√a² = √513.2751
a = 22.66

The second way is using the cosine ratio, adjacent/hypotenuse.  Side a is adjacent to the 25° angle, and the hypotenuse is 25:
cos 25 = a/25

Multiply both sides by 25:
25*cos 25 = (a/25)*25
25*cos 25 = a
22.66 = a

The third way is using the other angle.  First, find the measure of angle A by subtracting the other two angles from 180:
m∠A = 180-(90+25) = 180-115 = 65°

Side a is opposite ∠A; opposite/hypotenuse is the sine ratio:
a/25 = sin 65

Multiply both sides by 25:
(a/25)*25 = 25*sin 65
a = 25*sin 65
a = 22.66

#2)
A) Let side a be the one across from the 15° angle.  This would make the 15° angle ∠A.  We will define b as the side marked 4 and c as the side marked 3.  We will use the law of cosines:
a² = b²+c²-2bc cos A
a² = 4²+3²-2(4)(3)cos 15
a² = 16+9-24cos 15
a² = 25-24cos 15
a² = 1.82

Take the square root of both sides:
√a² = √1.82
a = 1.35

Use the law of sines to find m∠B:
sin A/a = sin B/b
sin 15/1.35 = sin B/4

Cross multiply:
4*sin 15 = 1.35*sin B

Divide both sides by 1.35:
(4*sin 15)/1.35 = (1.35*sin B)/1.35
(4*sin 15)/1.35 = sin B

Take the inverse sine of both sides:
sin⁻¹((4*sin 15)/1.35) = sin⁻¹(sin B)
50.07 = B

Subtract both known angles from 180 to find m∠C:
180-(15+50.07) = 180-65.07 = 114.93°

B)  Use the law of sines to find side b:
sin C/c = sin B/b
sin 52/12 = sin 45/b

Cross multiply:
b*sin 52 = 12*sin 45

Divide both sides by sin 52:
(b*sin 52)/(sin 52) = (12*sin 45)/(sin 52)
b = 10.77

Find m∠A by subtracting both known angles from 180:
180-(52+45) = 180-97 = 83°

Use the law of sines to find side a:
sin C/c = sin A/a
sin 52/12 = sin 83/a

Cross multiply:
a*sin 52 = 12*sin 83

Divide both sides by sin 52:
(a*sin 52)/(sin 52) = (12*sin 83)/(sin 52)
a = 15.11
3 0
3 years ago
Read 2 more answers
URGENT this is a question on my precalculus test I need help on. Thank you so much
sp2606 [1]

Answer:

a. 5i + 13j

b. 8i

c. 10i + 6j

Step-by-step explanation:

a. 2i + 6j + 3i + 7j = 5i + 13j

b. 3i + 7j + 5i - 7j = 8i

c. 2i + 6j + 3i + 7j + 5i - 7j = 10i + 6j

5 0
3 years ago
PLEASE HELP I WILL MARK BRAINLIST!!
marshall27 [118]

14: \frac{1}{2} = \frac{8}{1}    \frac{x}{y} = \frac{56}{1}

56/8 = 7 so multiply 1/2 by 7/1 to get the unidentified fraction "x/y"

1/2 · 7/1 = 7/2 or 3 1/2

15.  \frac{2}{5} = \frac{x}{100}

to get from 100 to 5, you would divide by 20, so to solve for x, multiply 2 by 20, to get a product of 40.

40/100 = 40% or D

16. \frac{3}{5} = \frac{12}{x}

A

7 0
2 years ago
A certain animal shelter has 28 cats and 46 dogs. how many dogs must be adopted and taken away from the animal shelter so that 7
postnew [5]
Make a box putt the 70% at the top on the ouside then sub 100%-70% the box should have 4 squares so then on the bottom of the box on the outside. it should be 30% then in the top left corner put 28 and under it 46 then do
46/28 × 70% over 1 see if that helps
3 0
3 years ago
Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))
Marrrta [24]

Answer:

\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

Step-by-step explanation:

\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}

Combine \frac{1}{c} + \frac{1}{h}

\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}

Combine the bottom, too.

=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}

Apply the fraction rule

=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}

Cancel

=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

Therefore, \frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

5 0
3 years ago
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