All categories

2 years ago

Please answer seriously will give brainliest

Mathematics

1 answer:

SashulF [63]2 years ago

7 0

Answer:

Step-by-step explanation:

33)

Chord - FE

Secant - GE

Diameter - FE

Radius - CG, CF , CE

Point of tangency - D

Chord - JL

Secant - MN

Diameter - JL

Radius - RK , JK , KL

Point of tangency - U

35)Chord - PN

Secant - LM

Diameter - PN

Radius - RQ, QP, QN

Point of tangency - K

You might be interested in

HELP NOW ASAP HURRY i don't got all day!!!!!!!!!!!!!

Anarel [89]

Answer:

15 3/4t^2 + 5 1/4

Explanation:

h(t)=-16t^2+400

We can add 400 then - 64 and show as 8^2

= h(t)=-16t^2+400 - 8^2

But we also need to change 16t^2;

We divide 64 into 16 = 16/64 =0.25

and show 16t^2 changes by 0.25 = 1/4

15 3/4t^2 + 400-64

Can now look like

15 3/4t^2 + 5 1/4

As 5 1/4 = 64 x 5 1/4 = 336

8 0

3 years ago

FAST!! Evaluate tan60/cos45 √6 √3/2 √2/3 1√6

evablogger [386]

Answer:

$\frac{\tan 60\degree}{\cos45 \degree}= \sqrt{6}$

Step-by-step explanation:

We want to evaluate

$\frac{\tan 60\degree}{\cos45 \degree}$

We use special angles or the unit circle to obtain;

$\frac{\tan 60\degree}{\cos45 \degree}=\frac{\sqrt{3}}{\frac{\sqrt{2}}{2}}$

This implies that;

$\frac{\tan 60\degree}{\cos45 \degree}=\sqrt{3}\div \frac{\sqrt{2}}{2}$

$\frac{\tan 60\degree}{\cos45 \degree}=\sqrt{3}\times \sqrt{2}$

$\frac{\tan 60\degree}{\cos45 \degree}= \sqrt{6}$

4 0

3 years ago

Read 2 more answers

Simplify the following radical square root of 45 times the power of 5

Alja [10]

The answer is 13584.1129633 which simplified is 13584 :)

7 0

3 years ago

Simplify:-6+{14+2[60-9(1+3)]}

yuradex [85]

1 + 3 = 4;
9 x 4 = 36;
60 - 36 = 24;
2 x 24 = 48;
14 + 48 = 62;
-6 + 62 = 56.

5 0

3 years ago

Read 2 more answers

What is the answer for number 8 and pls give step by step

Pie

Answer:

Trapezoid 1 (left side):

Base 1 = 2

Base 2 = 5

Trapezoid 2 (right side):

Base 1 = 6

Base 2 = 8

Step-by-step explanation:

1st trapezoid:

b_1 = x

b_2 = x + 3

h = 4

Hence, area (from formula) would be:

$A=\frac{h}{2}(b_1+b_2)\\A=\frac{4}{2}(x+x+3)\\A=2(2x+3)\\A=4x+6$

2nd trapezoid:

b_1 = 3x

b_2 = 4x

h = 2

Putting into formula, we get:

$A=\frac{h}{2}(b_1+b_2)\\A=\frac{2}{2}(3x+4x)\\A=1(7x)\\A=7x$

Let's equate both equations for area and find x first:

$4x+6=7x\\6=7x-4x\\6=3x\\x=\frac{6}{3}\\x=2$

We can plug in 2 into x and find length of each base of each trapezoid.

Trapezoid 1 (left side):

Base 1 = x = 2

Base 2 = x + 3 = 2 + 3 = 5

Trapezoid 2 (right side):

Base 1 = 3x = 3(2) = 6

Base 2 = 4x = 4(2) = 8

8 0

3 years ago