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1 year ago

Find the area of a semi circle with a diameter of 28cm

Mathematics

1 answer:

Alexandra [31]1 year ago

8 0

since it has a diameter of 28, then its radius must be half that or 14.

$\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=14 \end{cases}\implies A=\pi (14)^2\implies A=196\pi ~\hfill \stackrel{\stackrel{semi-circle}{half~that}}{98\pi }$

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Plz help!! Given: DE is a midsegment of ABC. Show all work.

topjm [15]

Answer:

AC=14

X=10

Y=1

Hoped this helped!

Step-by-step explanation:

DE is half of BC, therefore x=10. AE and EC have to be the same length, which is 7. SO, y=1 and x is 10.

5 0

2 years ago

Solve this system of equations by converting the equations into slope-intercept form. How many solutions does this system have?

Alex777 [14]

<h3>Answer: -3x-3y=-30= x=−y+10 </h3><h3>x+y+10= x=−y+10</h3>

8 0

2 years ago

Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

tigry1 [53]

Answer:

$P(A) = \frac{30}{100}$

$P(B) = \frac{77}{100}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

Step-by-step explanation:

Given

See attachment for proper format of table

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

$Yes = 22$ and $No = 8$

So, we have:

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

So, we have:

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

8 0

2 years ago

Solve the following quadratic equation 6x^2-5x-4=0

jeka94

Answer:

−

√

x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}

x=2a−b±b2−4ac

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

−

6x^{2}-5x-4=0

6x2−5x−4=0

a={\color{#c92786}{6}}

a=6

−

b={\color{#e8710a}{-5}}

b=−5

−

c={\color{#129eaf}{-4}}

c=−4

−

(

−

)

(

−

)

−

⋅

(

−

)

√

⋅

Step-by-step explanation:

6 0

2 years ago

How can you write the expression with rationalized denominator? sqrt3 - sqrt6 / sqrt 3 + sqrt 6

harina [27]

Answer: 2√2 - 3

Explanation:

The expession written properly is:

$\frac{ \sqrt{3}- \sqrt{6} }{ \sqrt{3}+ \sqrt{6} }$

To rationalize that kind of expressions, this is to eliminate the radicals on the denominator you use conjugate rationalization.

That is, you have to multiply both numerator and denominator times the conjugate of the denominator.

The conjugate of √3+√6 is √3 - √6, so let's do it:

$\frac{ \sqrt{3} - \sqrt{6} }{ \sqrt{3} + \sqrt{6} } . \frac{ \sqrt{3}- \sqrt{6} }{ \sqrt{3}- \sqrt{6} }$

To help you with the solution of that expression, I will show each part.

1) Numerator: (√3 - √6) . (√3 - √6) = (√3 - √6)^2 = (√3)^2 - 2√3√6 + (√6)^2 =

= 3 - 2√18 + 6 = 9 - 6√2.

2) Denominator: (√3 + √6).(√3 - √6) = (√3)^2 - (√6)^2 = 3 - 6 = - 3

3) Then the resulting expression is:

9 - 6√2
-----------
-3

Which can be further simplified, dividing by - 3

-3 + 2√2

Answer: 2√2 - 3

7 0

2 years ago

Read 2 more answers