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3 years ago

11. If 2 triangles have 2 of their angles

Mathematics

1 answer:

Shalnov [3]3 years ago

6 0

The answer is AA Similarity

You might be interested in

If f(x)=4x-11 what is the value of f(5)

Annette [7]

f(5) = 4(5) - 11

f(5) = 20 - 11

f(5) = 9

The answer is 9 :)

Please give BRAINIEST :)

Thanks

5 0

3 years ago

Read 2 more answers

Giải dumfmmmm voqisiiiiiiiiii

Brums [2.3K]

Answer:

私はあなたの言語を理解していなかった

Step-by-step explanation:

お邪魔して申し訳ありません

ところで、あなたが与えたポイントに感謝しますあなたは、あなたが、あなたのおかげで、あなたは、このビデオをありがとう

8 0

2 years ago

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found:

$\alpha_{1} = -2\cdot \alpha_{3}$

$\alpha_{1} = -2\cdot k$

$\alpha_{2}= -2\cdot \alpha_{3}$

$\alpha_{2} = -3\cdot k$

It is evident that $\alpha_{1}$ and $\alpha_{2}$ are multiples of $\alpha_{3}$ , which means that the set of vector are linearly dependent.

$p_{1}(t) = 1$ , $p_{2}(t)=t^{2}$ and $p_{3}(t) = 3+3\cdot t +t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2}+ \alpha_{3}\cdot (3+3\cdot t+t^{2}) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1+(\alpha_{2}+\alpha_{3})\cdot t^{2}+3\cdot \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1}+3\cdot \alpha_{3} = 0$

$\alpha_{2} + \alpha_{3} = 0$

$3\cdot \alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

The set of vectors A and C are linearly independent.

4 0

3 years ago

Anson estimated he would need 17 hours to complete his science project. He actually needed 19 1 2 hours. What was his percent er

STatiana [176]

Answer:

The percentage of error is 14.71 %.

Step-by-step explanation:

Given:

Estimated time for completion of project is, $ET=17\ h$

Actual time for completion of project is, $AT=19\frac{1}{2}=\frac{39}{2}\ h$

Difference in error = $AT - ET = \frac{39}{2}-17=\frac{39-34}{2}=\frac{5}{2}=2.5\ h$

Percentage in error = $\frac{2.5}{17}\times 100=14.71\%$

Therefore, the percentage of error is 14.71 %

7 0

3 years ago

Need help on both of these Calculus AB mock exams.

babymother [125]

AB1

(a) $g$ is continuous at $x=3$ if the two-sided limit exists. By definition of $g$ , we have $g(3)=f(3^2-5)=f(4)$ . We need to have $f(4)=1$ , since continuity means

$\displaystyle\lim_{x\to3^-}g(x)=\lim_{x\to3}f(x^2-5)=f(4)$

$\displaystyle\lim_{x\to3^+}g(x)=\lim_{x\to3}5-2x=-1$

By the fundamental theorem of calculus (FTC), we have

$f(6)=f(4)+\displaystyle\int_4^6f'(x)\,\mathrm dx$

We know $f(6)=-4$ , and the graph tells us the *signed* area under the curve from 4 to 6 is -3. So we have

$-4=f(4)-3\implies f(4)=-1$

and hence $g$ is continuous at $x=3$ .

(b) Judging by the graph of $f'$ , we know $f$ has local extrema when $x=0,4,6$ . In particular, there is a local maximum when $x=4$ , and from part (a) we know $f(4)=-1$ .

For $x>6$ , we see that $f'\ge0$ , meaning $f$ is strictly non-decreasing on (6, 10). By the FTC, we find

$f(10)=f(6)+\displaystyle\int_6^{10}f'(x)\,\mathrm dx=7$

which is larger than -1, so $f$ attains an absolute maximum at the point (10, 7).

$k(3)=\dfrac{\displaystyle3\int_4^9f'(t)\,\mathrm dt-12}{3e^{2f(3)+5}-3}$

From the graph of $f'$ , we know the value of the integral is -3 + 7 = 4, so the numerator reduces to 0. In order to apply L'Hopital's rule, we need to have the denominator also reduce to 0. This happens if

$3e^{2f(3)+5}-3=0\implies e^{2f(3)+5}=1\implies 2f(3)+5=\ln1=0\implies f(3)=-\dfrac52$

Next, applying L'Hopital's rule to the limit gives

$\displaystyle\lim_{x\to3}k(x)=\lim_{x\to3}\frac{9f'(3x)-4}{6f'(x)e^{2f(x)+5}-1}=\frac{9f'(9)-4}{6f'(3)e^{2f(3)+5}-1}=-\dfrac4{21e^{2f(3)+5}-1}$

which follows from the fact that $f'(9)=0$ and $f'(3)=3.5$ .

Then using the value of $f(3)$ we found earlier, we end up with

$\displaystyle\lim_{x\to3}k(x)=-\dfrac4{20}=-\dfrac15$

(d) Differentiate both sides of the given differential equation to get

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=2f''(2x)(3y-1)+f'(2x)\left(3\dfrac{\mathrm dy}{\mathrm dx}\right)=[2f''(2x)+3f'(2x)^2](3y-1)$

The concavity of the graph of $y=h(x)$ at the point (1, 2) is determined by the sign of the second derivative. For $y>\frac13$ , we have $3y-1>0$ , so the sign is entirely determined by $2f''(2x)+3f'(2x)^2$ .

When $x=1$ , this is equal to

$2f''(2)+3f'(2)^2$

We know $f''(2)=0$ , since $f'$ has a horizontal tangent at $x=2$ , and from the graph of $f'$ we also know $f'(2)=5$ . So we find $h''(1)=75>0$ , which means $h$ is concave upward at the point (1, 2).

(e) $h$ is increasing when $h'>0$ . The sign of $h'$ is determined by $f'(2x)$ . We're interested in the interval $-1$ , and the behavior of $h$ on this interval depends on the behavior of $f'$ on $-2$ .