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2 years ago

Which correctly shows 5.05 and 5.5

Mathematics

2 answers:

elena-14-01-66 [18.8K]2 years ago

7 0

Answer:

how are we gonna answer the question if there is no picture of it

Step-by-step explanation:

yuradex [85]2 years ago

4 0

Answer:

You did not attach anything. Please make sure to attach your questions so you can get all the help you need!!!

Step-by-step explanation:

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How are quadratic equations different from linear equations?

worty [1.4K]

The fundamental difference between the linear and quadratic functions is observed in their respective graphs. A linear function is represented by a straight line and a quadratic function is represented by a curve called a parabola.

See file attached.

3 0

3 years ago

Jenin andrea marie are picking apples andre pics three times as many pounds as maria jane picks two times as many pounds is andr

Sindrei [870]

<span>Let x=apples that Andre picks
Then x/3=apples that Maria picks
And 2x=apples that Jane picks
Soooo,
x +(x/3)+2x=840 multiply each term by 3

3x+x+6x=2520 collect like terms
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5 0

3 years ago

Write the equation of a line that is parallel to 2y=x-10 passing through (2,1)

MariettaO [177]

the line that is parallel will be y = 1/2x

6 0

3 years ago

Read 2 more answers

Calculate the area of triangle ABC with altitude AD, given A(-5, 4), B(-6, 1), C(0, 1), and D(-5, 1).

snow_lady [41]

Answer:

the area of triangle ABC = 9

Step-by-step explanation:

Check the photo below for the details.

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