Evaluate bcl, given a = 5, b=-3, and c= -2.

If the sin of angle x is 4 over 5 and the triangle was dilated to be two times as big as the original, what would be the value o

beks73 [17]

Answer:

Sin of x does not change

Step-by-step explanation:

Whenever a triangle is dilated, the angle remains the same as well as the ratio for sides of triangle. For smshapes with dimensions, when shapes are dilated the dimensions has increment with common factor.

From trigonometry,

Sin(x)=opposite/hypotenose

Where x=4/5

Sin(4/5)= opposite/hypotenose

But we were given the scale factor of 2 which means the dilation is to two times big.

Then we have

Sin(x)=(2×opposite)/(2×hypotenose)

Then,if we divide by 2 the numerator and denominator we still have

Sin(x)=opposite/hypotenose

Which means the two in numerator and denominator is cancelled out.

Then we still have the same sin of x. as sin(4/5)

Hence,Sin of x does not change

4 0

3 years ago

Which of the following graphs could represent a 6th-degree polynomial function, with 3 distinct zeros, 1 zero with a multiplicit

klasskru [66]

Answer:

a=1

b=5

c=1

d=3

1+5+1+3 = 10

5 0

3 years ago

Someone please help I will give brainliest to the correct answer

USPshnik [31]

Answer:

X inter= (-1.375,0)

Y inter= (0,2.2)

Step-by-step explanation:

I used "desmos graphing calculator"

If answer is right, check out desmos graphing calculator. It is a really good website where you enter equation and it graphs it for you.

I use desmos for questions that have anything to do with graphs

4 0

3 years ago

Read 2 more answers

For p = $3000, r = 3.5%,and t = 5 years, Find the balance in an account When interest is compounded (a) quarterly, (b) monthly,a

MAVERICK [17]

Answer:

a) $3,571.02

b) $3,572.9

c) $3,573.74

Step-by-step explanation:

Data provided in the question:

p = $3000,

r = 3.5%,

t = 5 years

a) quarterly

number of periods in a year, n = 4

Interest rate per period = 3.5% ÷ 4 = 0.875%

Now,

$A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}$

A = total amount

n = number of times compounded per year

on substituting the respective values, we get

A = 3000 × $\left( 1 + \frac{ 0.035 }{ 4 } \right)^{\Large{ 4 \cdot 5 }}$

A = 3000 × [/tex]\cdot { 1.00875 } ^ { 20 }[/tex]

A = 3000 × 1.19034

A = $3,571.02

b) monthly

number of periods in a year, n = 12

Now,

A = $p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}$

on substituting the respective values, we get

A = 3000 × $\left( 1 + \frac{ 0.035 }{ 12 } \right)^{\Large{ 12 \cdot 5 }}$

A = 3000 × [/tex]\cdot { 1.002917} ^ { 60 }[/tex]

A = 3000 × 1.190967

A = $3,572.9

c) continuously

A = $pe^{r\times t}$

on substituting the respective values, we get

A = 3,000 × $e^{0.035\times 5}$

or

A = 3,000 × $e^{0.175}$

or

A = 3,000 × 1.1912

or

A = $3,573.74

5 0

3 years ago

Read 2 more answers

Find the length of the third side to the nearest tenth.<br> 8<br> 5

brilliants [131]

Answer:

The length of third side is: 6.3

Step-by-step explanation:

Given triangle is a right triangle where

Base = ?

Perpendicular = 5

Hypotenuse = 8

Let x be the base

As it is a right-angled triangle, Pythagoras theorem can be used to find the third side

$Hypotenuse^2 = Base^2 + Perpendicular^2$

Putting the known values

$(8)^2 = x^2 + (5)^2\\x^2 = (8)^2-(5)^2\\x^2 = 64-25\\x^2 = 39\\$

Taking square root on both sides

$\sqrt{x^2} = \sqrt{39}\\x = 6.24499$

Rounding off to the nearest 10th

Base = 6.3

Hence,

The length of third side is: 6.3

6 0

3 years ago