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3 years ago

Can I get help with this? thanks.

Mathematics

1 answer:

Basile [38]3 years ago

6 0

The range is the value of y. We know that Ix-12I is always ≥0, no matter what x is, so Ix-12I-2 is always ≥ -2, the answer is B.

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Thank you!!!!!!!!!!!!

AlekseyPX

Answer:

Step-by-step explanation:

square root of 36 is 6 ^w^

u = 6

5 0

3 years ago

7 copies of the sum of 8 fifths and 4

Agata [3.3K]

Answer:

39 1/5 or 39.2

Step-by-step explanation:

<u>Sum of 8 fifths and 4:</u>

8/5 + 4 =
1 3/5 + 4 =
5 3/5

<u>7 copies of the of sum of 8 fifths and 4:</u>

7 × 5 3/5 =
7( 5 + 3/5) =
35 + 7(3/5) =
35 + 21/5 =
35 + 4 1/5 =
39 1/5 or 39.2

4 0

3 years ago

What the correct answer do not want the wrong answer please

Crank

Answer:

388.5yd²

Step-by-step explanation:

We have Triangle TUV

In the question, we are given already

Angle U = 32°

Angle T = 38°

Angle V = ???

Side t = 31yd

Side u = ?

Side v = ?

Area of the triangle= ?

Step 1

We find the third angle = Angle V

Sum of angles in a triangle = 180°

Third angle = Angle V = 180° - (32 + 38)°

= 180° - 70°

Angle V = 110°

Step 2

Find the sides u and v

We find these sides using the sine rule

Sine rule or Rule of Sines =

a/ sin A = b/ Sin B

Hence for triangle TUV

t/ sin T = u/ sin U = v/ sin V

We have the following values

Angle T = 38°

Angle U = 32°

Angle V = 110°

We are given side t = 31y

Finding side u

u/ sin U= t/ sin T

u/sin 32 = 31/sin 38

Cross Multiply

sin 32 × 31 = u × sin 38

u = sin 32 × 31/sin 38

u = 26.68268yd

u = 26.68yd

Finding side x

v / sin V= t/ sin T

v/ sin 110 = 31/sin 38

Cross Multiply

sin 110 × 31 = v × sin 38

v = sin 110 × 31/sin 38

v = 47.31573yd

v = 47.32yd

To find the area of triangle TUV

We use heron formula

= √s(s - t) (s - u) (s - v)

Where S = t + u + v/ 2

s = (31 + 26.68 + 47.32)/2

s = 52.5

Area of the triangle = √52.5× (52.5 - 31) × (52.5 - 26.68 ) × (52.5 - 47.32)

Area of the triangle = √150967.6032

Area of the triangle = 388.5454973359yd²

Approximately to the nearest tenth =388.5yd²

5 0

3 years ago

Marie has 50 pages in her notebook. She made a graph of the kinds of writing on all of the pages she has

MissTica

Answer:

Step-by-step explanation:

50 - 2-3-6-3-5 = 31

4 0

3 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago