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3 years ago

PLEASE HELP ME I REALLY NEED HELP

Mathematics

1 answer:

Dominik [7]3 years ago

4 0

Answer:

Step-by-step explanation:

We expect you're supposed to answer, NO. This is because an angle and two sides are specified, where the angle is not between the two sides. There is no SSA theorem for congruence, so this specification is generally considered to be insufficient.

_____

Additional comment

The given information is enough to completely specify the given triangle. The reason this is true is that the given angle is opposite the longest of the given sides. A general SSA theorem cannot guarantee that will be the case, so there is no general SSA congruence theorem.

Provided that you have corresponding side and angle information for the other triangle, congruence can be determined (yes). A single triangle by itself cannot be shown congruent to anything.

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Find the surface area of the square pyramid if the slant height is 13 mm.

Tanzania [10]

Simple Question:

Find the surface area of the square pyramid if the slant height is 13 mm.

Options:

A. 264 mm

B. 288 mm

C. 335 5 mm

D. 407 mm

Answer:

A) 264mm

Step-by-step explanation:

11 x 13 = 143 = 264mm

5 0

4 years ago

PLEAAEEEEEE HELP! CORRECT ANSWER GETS BRAINLIEST URGET!!

Georgia [21]

Answer:

Graph F

Step-by-step explanation:

Given

x + 3 < 7 ( subtract 3 from both sides )

x < 4

Since x is less than 4 then this is indicated by an open circle at 4 on the number line with the arrow pointing to the left as shown on graph F

4 0

4 years ago

Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
$=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2$
$=\dfrac{\ln5}4$

Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

7 0

3 years ago

A mine produces 42.5 tons of coal per hour. How much coal will the mine produce in 12.5 hours?

bonufazy [111]

Answer:

531.25 tons per 12.5 hr

Step-by-step explanation:

42.5 per hr

× 12.5

531.25 tons per 12.5 hrs

7 0

4 years ago

I need help finding the perimeter and area

Murrr4er [49]

Answer:

A = 658

P = 110

Step-by-step explanation:

Split the shape into 2 rectangles and calculate the missing side measures.

Add up all the side measures to get the perimeter.

Add up the areas of the 2 rectangles to get the total area

5 0

4 years ago