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Dafna11 [192]
3 years ago
8

Given the equation 4x2 − 8x + 20 = 0, what are the values of h and k when the equation is written in vertex form a(x − h)2 + k =

0?
h = 4, k = −16
h = 4, k = −1
h = 1, k = −24
h = 1, k = 16
Mathematics
1 answer:
GenaCL600 [577]3 years ago
8 0

Answer:

The correct option is:

h = 1, k = 16

Step-by-step explanation:

y=4x^2-8x+20 =0

It is a quadratic formula in standard form:

ax^2+bx+c

where a = 4 , b = -8 and c=20

The vertex form is:

a(x − h)2 + k = 0

h is the axis of symmetry and (h,k) is the vertex.

Calculate h according to the following formula:

h = -b/2a

h= -(-8)/2(4)

h = 8/8

h = 1

Substitute k for y and insert the value of h for x in the standard form:

ax^2+bx+c

k = 4(1)^2+(-8)(1)+20

k = 4-8+20

k=-4+20

k = 16

Thus the correct option is h=1, k=16....

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Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\

With respect to reference angle A, we have:

  • opposite side = a = 4
  • adjacent side = b = \sqrt{33}
  • hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\

Then cosine which is adjacent over hypotenuse

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\

Tangent is the ratio of opposite over adjacent

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

  • cosecant, abbreviated as csc, is the reciprocal of sine
  • secant, abbreviated as sec, is the reciprocal of cosine
  • cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{  ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

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Summary:

The missing side is b = \sqrt{33}

The 6 trig functions have these results

\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

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Ashley mixes two types of soft drinks with different types of concentration: one soft drink has 20% sugar and the other drink ha
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Answer:

Sugar concentration of the mixed soft drink will be <em>32.5%.</em>

<em></em>

Step-by-step explanation:

Sugar concentration of first drink = 20%

Volume of first soda = 250 ml

Sugar in first soda = 20% of 250 ...... (1)

Sugar concentration of second drink = 45%

Volume of second soda = 250 ml

Sugar in second soda = 45% of 250 ...... (2)

Let x\% be the sugar concentration in resultant mixture.

Volume of mixture = 250 + 250 = 500 ml

Sugar in mixture = x\% of 500 ml ...... (3)

Total sugar concentration (Adding (1) and (2)) and equating it to equation (3):

20% of 250 + 45% of 250 = x\% of 500

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Hence, Sugar concentration of the mixed soft drink will be <em>32.5%.</em>

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