Please help ASAP I need this question answered right now please help if you can thanks to anyone that does

Ket [755]

Answer: BD = 9 units

Step-by-step explanation:

ABQ and CDQ are congruent triangles. That means their angle measures are the same but their lengths may be proportional.

19.2 / 32 = 0.6

15 * 0.6 = 9

So the length of side BD is 9 units.

4 0

3 years ago

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What is a GPA? How is it calculated and determined? If a person is required to have a GPA 95 or higher, what does that mean?

tester [92]

Answer:

GPA stands for grade point average. the highest you can get is 4.5. There is no such things as a 95 GPA. To calculate, take the number or point per grade you have, A is 4.0, B is 3.0, C is 2.0, and so on, and add them all up and then divide by the number of classes you are taking.

Step-by-step explanation:

4 0

3 years ago

(HELP ASAAPPP!!!!) A restaurant staff made 54 sandwiches in 18 minutes. At that rate, how many sandwiches would the restaurant s

KatRina [158]

75 бутеров!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

4 0

2 years ago

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Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago

Angie baked 100 cookies and 20 brownies. She wants to split them into equal groups for the big sale. Each group must have the

Tresset [83]

Stop copying the question, it's confusing

so 100 cookies and 20 brownies
what is the greatest number you can divide them both by?

find the GCF
factor them
100=2*2*5*5
20=2*2*5
so the GCF is the common group to both or 2*2*5 or 20

100/20=5
20/20=1

there are 20 groups, each wit 5 cookies and 1 brownie

20 groups

7 0

2 years ago