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3 years ago

Determine the height of each triangle. round to the nearest foot.

Mathematics

2 answers:

-BARSIC- [3]3 years ago

7 0

Answer:

The answer is 7 ft

Step-by-step explanation:

statuscvo [17]3 years ago

5 0

Do you have photos for the triangle?

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Which of these is an example of the Commutative property of addition? And Explain why? 3+5=4+4 or 3+5=5+3

arlik [135]

Answer:

3+5=5+3

Step-by-step explanation:

a+b=b+a

comunitive property

3 0

3 years ago

Read 2 more answers

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

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

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

3 0

3 years ago

9(8s - 5f) use f = 3 and s = 4

Savatey [412]

Answer:

153

Step-by-step explanation:

8x4=32

5x3=15

32-15=17

9x17=153

5 0

3 years ago

Read 2 more answers

Simplify 3(2f+6)+7 thanks

Naya [18.7K]

{Expand} 3(2f+6): 6f+18

6f+18+7

18+7=25

= 6f+25

4 0

3 years ago

In the year 2003, a company made $6.8 million in profit. For each consecutive year after that, their profit increased by 13%. Ho

LenKa [72]

Answer:

$7.7 million

Step-by-step explanation:

Given: $6.8 million profit made by company.

13% increase in profit every year.

Now, finding the company´s profit in the year of 2006.

As we know there is 13% increase in profit.

∴ Profit increase= $\frac{13}{100} \times 6.8= \$ 0.884$

Next adding company´s profit with increase in profit to get profit in the year 2006.

∴ $6.8+0.884= \$ 7.684$ ≅ $7.7

The company´s profit in the year 2006 is $7.7 million.

6 0

3 years ago