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2 years ago

I need help thanks hullo

Mathematics

2 answers:

babymother [125]2 years ago

8 0

The answer is b for your answer
But we can put it that way

Novay_Z [31]2 years ago

3 0

Answer:

the answer would be "A"

Step-by-step explanation:

hope this helpssss

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<img src="https://tex.z-dn.net/?f=%24%5Coverline%7BBY%7D%24" id="TexFormula1" title="$\overline{BY}$" alt="$\overline{BY}$" alig

erica [24]

Answer:

BZ = 64/23 = 2 18/23

Step-by-step explanation:

The angle bisectors divide the triangle into proportional parts, so we have ...

CB/CY = AB/AY

CB = CY(AB/AY) = 4(8/6) = 16/3 . . . . multiply by CY and evaluate

and ...

BZ/CB = AZ/AC

BZ = AZ(CB/AC) . . . . multiply by CB

But we have ...

AZ = 8 -BZ, so this becomes ...

BZ = (8 -BZ)(16/3)/(10)

BZ(1 +8/15) = 64/15 . . . . add the BZ term on the right to both sides

BZ = (64/15)(15/23) = 64/23 . . . . divide by the coefficient of BZ

BZ = 64/23 = 2 18/23

6 0

2 years ago

The sum of 1/7 + 2/5

matrenka [14]

Answer:

19/35

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Grumpy Corp drug tests all of the recent college graduates it hires each year. The drug test currently used correctly determines

Vedmedyk [2.9K]

Answer:

Hence, the probability of a Negative test is $P(T^C)=0.5904$ .

Step-by-step explanation:

(a)

Let $D=$ Drug user and $T=$ Test positive

Probability of drug users $P(D)=36\%=0.36$

$P(\frac{T}{D})=96\%=0.96$

$P(\frac{T^c}{D^c})=90\%=0.90$

(b)

Find the Probability of a Negative test,

$P(T^C)=P(\frac{T^C}{D^C})\cdot P(D^C)+P(\frac{T^C}{D})\cdot P(D)$

$\Rightarrow P(T^C)=0.9\cdot(1-0.36)+(1-0.96)\cdot 0.36$

$\Rightarrow P(T^C)=0.9\cdot 0.64+0.04\cdot 0.36$

$\Rightarrow P(T^C)=0.576+0.0144$

$\Rightarrow P(T^C)=0.5904$

4 0

2 years ago

Which one of the following equations represents a circle with its center of origin?

babymother [125]

Answer:

A) x² + y² = 5

Step-by-step explanation:

<u>The equation of the circle:</u>

(x - h)² + (y - k)² = r²

The center is (h, k)

<u>Wen the center is the origin we have, h = 0, k = 0:</u>

x² + y² = r²

The only equation in same format is the first one

3 0

2 years ago

ſ x - 5y = -9( 4x + 4y = - 12Step 1 of 2: Determine if the point (-4,1) lies on both of the lines in the system of equations by

Phoenix [80]

Given:

$\begin{gathered} x-5y=-9\ldots\ldots(1) \\ 4x+4y=-12\ldots\text{.}\mathrm{}(2) \end{gathered}$

The given point is (-4,1)

Let's check the ordered pair (-4,1) in the first equation

$\begin{gathered} x-5y=-9 \\ (-4)-5(1)=-9 \\ -4-5=-9 \\ -9=-9 \end{gathered}$

Hence, (-4,1) is a solution of the first equation x-5y=-9.

Now, let's check (-4,1) in the second equation.

$\begin{gathered} 4x+4y=-12 \\ 4(-4)+4(1)=-12 \\ -16+4=-12 \\ -12=-12 \end{gathered}$

So, (-4,1) is a solution of the second equation 4x+4y=-12.

Hence, (-4,1) is a solution of both equation in the system, then it is a solution to the overall system.

3 0

4 months ago