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3 years ago

Will mark brainliest

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

5 0

Answer:

Yes and SSS

Step-by-step explanation:

Let’s write down the ratios to figure out the first question:

RP:MN=5:3

NP:ML=10:6

RN:LN=15:9

Notice that we can simplify all of those ratios to 5:3. That means that the sides are all similar, so yes, the triangles are similar.

For the second question: They are similar by because all sides are similar in the same ratio. That means that they are similar by the SSS similarity.

Hope that helps

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3x+10y=13 4(-2y+x)-9x=13

mart [117]

Answer:

$x=-9$

$y=4$

Step-by-step explanation:

$3x+10y=13$ ← Equation 1

$4(-2y+x)-9x=13\\$

By simplifying the above equation we get,

$-8y+=4x-9x=13$ (Simplifying the brackets)

$-8y-5x=13$ (By subtracting $-9x$ from $+4x$ ) ← Equation 2

Multiply Equation 1 by 5 and Equation 2 by 3 and add them together.

Equation 1 multiplied by 5 will give,

$15x+50y=65$

Equation 2 multiplied by 3 will give,

$-24y-15x=39$

Add those together,

$15x+50y-24y-15x=65+39$

$26y=104$ (After simplifying $x$ values)

Therefor $y=4$

By substituting $y=4$ to Equation 1 we get,

$3x+10*4=13$ $3x+40=13\\3x=-27\\x=-9$

Therefor we can say,

$x=-9$

$y=4$

8 0

4 years ago

The quotient of n and 7 is 39

Scorpion4ik [409]

Let n = the unknown number.

Quotient tells you use division.

n/7 = 39

Multiply both sides by 7 to cancel out the number in the denominator and isolate the variable.

n = 39(7) = 273

n = 273

4 0

3 years ago

Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +

slava [35]

$\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx$

Taking $x=7\tan\theta$ gives $\mathrm dx=7\sec^2\theta\,\mathrm d\theta$ , so that the integral becomes

$\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta$
$=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta$

When $\sec\theta>0$ , we have

$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta$
$=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta$

and from here we can substitute $u=\tan\theta$ to proceed from here.

Quick note: When we set $x=7\tan\theta$ , we are implicitly enforcing $-\dfrac\pi2$ just so that the substitution can be undone later via $\theta=\tan^{-1}\dfrac x7$ . But note that over this domain, we automatically guarantee that $\sec\theta>0$ , so the absolute value bars can be dropped immediately.

6 0

3 years ago

Which equation is represented by the phrase “one-fourth of a number, increased by eight equals sixteen”?

Alexandra [31]

(1/4)x + 8 = 16

or...
(x/4) + 8 = 16

7 0

4 years ago

Read 2 more answers

Pls help reward brainiest

slega [8]

Answer:

(5, 2) & (-4, -9)