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2 years ago

Given Isosceles Trapezoid TRAP, if mP=98, find mT

Mathematics

1 answer:

VikaD [51]2 years ago

5 0

Answer:

∠ T = 82°

Step-by-step explanation:

In a Isosceles trapezoid, any lower base angle is supplementary to any upper base angle , so

∠ T + ∠ P = 180°

∠ T + 98° = 180° ( subtract 98° from both sides )

∠ T = 82°

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One die is rolled 3 times, what is the chance of getting all 4's?

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6^3. 216. When using chance, take the possible outcomes(six for a die) and raise it to the power of the number of repetitions.

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3 years ago

The points (8,18) and (24,54) form a proportional relationship. Find the slope of the line through the

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Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

<em>Hence the limit of the function </em> $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

3 0

3 years ago

Stan has made a $125.30 monthly deposit into an account that pays 1.5% interest, compounded monthly, for 35 years. he would now

Andrew [12]

The annuity of the monthly deposit into an account that pays 1.5% interest, compounded monthly, for 35 years is $333.71

<h3>What is annuity?</h3>

An annuity is a series of payments made at equal period of time.

future value = annuity x [(1 + i)ⁿ - 1] / i

annuity = $125.30

i = 1.5% / 12 = 0.00125

n = 35 years x 12 months = 420

future value = $125.30 x [(1 + 0.00125)⁴²⁰ - 1] / 0.00125

future value = $69,156.049 ≈ $69,156.05

annuity = [i x (present value)] / [1 - (1 + i)⁻ⁿ]

i = 1.5% / 12 = 0.00125

n = 20 years x 12 months = 240

present value = $69,156.05

annuity = (0.00125 x $69,156.05) / [1 - (1 + 0.00125)⁻²⁴⁰]

annuity = $86.45 / 0.25904

= $333.71

Learn more about annuity;

brainly.com/question/23554766

4 0

1 year ago