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3 years ago

Help plz giving out brianlest and 67 points please help

Mathematics

1 answer:

Bezzdna [24]3 years ago

5 0

Its the top answer because 3 x 1/4 is 3/4 which is the a value in function 2

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I asked my dad how to solve this question and he part helped me and I was like ohhhhhhhhhhh so now I want you to solve it. I'll

matrenka [14]

Answer:

Whats the question? I can surely answer it.

3 0

3 years ago

60 - 3d + 20 - 6 = 4d

iragen [17]

60-3d + 20-6 =4d
74-3d =4d
-3d =4d -74
-7d = -74
D=74/7

4 0

3 years ago

Read 2 more answers

Find the area of quadrilateral ABCD

andreev551 [17]

Answer:

A ≈ 28.5

Step-by-step explanation:

a, b, c

P = a + b + c

Semiperimeter s = $\frac{a+b+c}{2}$

A = $\sqrt{s(s-a)(s-b)(s-c)}$

~~~~~~~~~~~~~~~

$P_{ABC}$ = 4.3 + 2.89 + 6.81 = 14

s = 14 ÷ 2 = 7

$A_{ABC}$ = $\sqrt{7(7-4.3)(7-2.89)(7-6.81)}$ = √14.75901 ≈ 3.84

$P_{BCD}$ = 8.59 + 7.58 + 6.81 = 22.98

s = 22.98 ÷ 2 = 11.49

$A_{BCD}$ = $\sqrt{11.49(11.49-8.59)(11.49-7.58)(11.49-6.81)}$ = √609.7343148 ≈ 24.6928

$A_{ABCD}$ = 3.84 + 24.6928 ≈ 28.5

3 0

3 years ago

1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β

slava [35]

Step-by-step explanation:

$\tan \beta = \frac{1}{ \cot \beta } = - \frac{3}{2 \sqrt{10} } = - \frac{3 \sqrt{10} }{20}$

$\csc \beta \tan \beta = \frac{1}{ \cos \beta } = \sec \beta$

Therefore,

$\sec \beta = ( \frac{7}{3} )( - \frac{3 \sqrt{10} }{20} ) = - \frac{7 \sqrt{10} }{20}$

$\csc y = \frac{1}{ \sin y} = - \frac{ \sqrt{6} }{2}$

$= > \sin y = - \frac{ \sqrt{6} }{3}$

Use the identity

$\cos y = \sqrt{1 - \sin ^{2} y} \: \: \: \: \: \: \: \: \: \: \: \\ = \sqrt{1 - {( - \frac{ \sqrt{6} }{3}) }^{2} } = - \frac{ \sqrt{3} }{3}$

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

$\tan \beta = \frac{ \sin y}{ \cos y} =( - \frac{ \sqrt{6} }{3} )( - \frac{3}{ \sqrt{3} } ) = \sqrt{2}$

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

$\sin x = \sqrt{1 - \cos ^{2} x} = \sqrt{ \frac{11}{6} }$

Solving for tan x:

$\tan x = \frac{ \sin x}{ \cos x} = (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) = \frac{ \sqrt{66} }{5}$

5 0

3 years ago

Try to prove the hypothesis from part B that the sum of a rational and an irrational number is an irrational number.

Maurinko [17]

The next step of your proof is to subtract (a/b) from both sides.

Then you get, x = (m/n) - (a/b)

Since rationals are closed over addition, (m/n) + (-a/b) is a rational number.

Therefore, x (an irrational number) = a rational number This is a false statement which is a contradiction. So, the assumption was incorrect.

Thus, the sum of a rational and irrational number is an irrational number. QED

8 0

3 years ago