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2 years ago

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If ray BD bisects angle CBE, ray BC is transparent to ray BA, angle CBD = (3x + 25) degrees, and angle DBE = (7x - 19) degrees,

find angle ABD

2 answers:

Alex777 [14]2 years ago

7 0

Check the picture below.

Alenkasestr [34]2 years ago

5 0

Answer: The required measure of angle ABD is 148°.

Step-by-step explanation: Given that ray BD bisects angle CBE, ray BC is transparent to ray BA.

Also,

$m\angle CBD=(3x+25)^\circ,~~m\angle DBE=(7x-19)^\circ,~~m\angle ABC=90^\circ.$

We are to find the measure of angle ABD.

Since ray BD bisects angle CBE, so we have

$m\angle CBD=m\angle DBE\\\\\Rightarrow (3x+25)^\circ=(7x-19)^\circ\\\\\Rightarrow 3x+25=7x-19\\\\\Rightarrow 7x-3x=25+19\\\\\Rightarrow 4x=44\\\\\Rightarrow x=\dfrac{44}{4}\\\\\Rightarrow x=11.$

So, we get

$m\angle CBD=(3\times11+25)^\circ=58^\circ.$

Therefore,

$m\angle ABD=m\angle ABC+m\angle CBD=90^\circ+58^\circ=148^\circ.$

Thus, the required measure of angle ABD is 148°.

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T= 300 (d−15) 2 +20space, T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, start superscript, 2,

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As can be read from your statement written "T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, strt superscript, 2, end superscript, divided by, 300, end fraction, plus, 20", I hope your model equation is this :

$T = \frac{(d-15)^{2}}{300} + 20$

Hope this is your question, if not I think you will, still be able to find an answer of your question based on this solution

As we have to find lowest average temperature, So for minimum of a function its derivative is equal to 0 there.

So lets find derivative of T function first

So first expand $(d-15)^{2}$ as (d-15)(d-15)

we will use FOIL to multiply these

so $(d-15)(d-15) = d^{2} -15d -15d +225$

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so we have $T = \frac{d^{2} -30d +225}{300} +20$

Now we will derivate each term here,300 in denominator is constant so that will come as it in in denominator.

To derivate terms in $dx^{2} -30d +225$ we will use power rule formula:

$(x^{n} )'= nx^{n-1}$

so derivative of $(d^{2} )'= 2d^{2-1} = 2d^{1} = 2d$

Then derivative of d will be 1

so that of -30d will be -30

then derivate of constant -225 will be 0

so we will have derivative as $\frac{2d-30}{300}$ for the fraction part and then derivative of +20 is again 0 as its constant term

$T' = \frac{2d-30}{300}$

For minimum we will put this derivative =0

$0 = \frac{2d-30}{300}$

Now solve for d

times both sides by 300

$0 \times 300 = \frac{2d-30}{300} \times 300$

$0 = 2d-30$

$0 +30 = 2d -30 +30$

$30 = 2d$

$\frac{30}{2} = \frac{2d}{2}$

$15 = d$

So now we have to find value of lowest temperature.

For that simply plug 15 in d place in original T function equation

$T = \frac{(d-15)^{2}}{300} + 20$

$T = \frac{(15-15)^{2}}{300} + 20$

$T = 20$

So T = 20 °C is the lowest average temperature and the answer.

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