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3 years ago

What else would need to be congruent to show that triangle ABC and triangle XYZ is congruent by AAS

Mathematics

1 answer:

julia-pushkina [17]3 years ago

7 0

Answer:

C. ∠B ≅ ∠Y

Step-by-step explanation:

This is because AAS means 2 angles and 1 side are congruent.

Which means there are two possible answers: B or C. But answer B says angle C and Z are congruent which would make both triangles congruent by ASA and not AAS

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What is 16,473 to the nearest hundred

Verizon [17]

The number "4" is in the hundreds place. To the right is 7, and since it's greater than five it rounds up. So 16,473 to the nearest hundred is 16,500. Hope this helped!

4 0

3 years ago

PLZ help me fast 2 mins left

jek_recluse [69]

Answer:

$\sqrt{-5}=i\sqrt{5}$

$\sqrt{-25}=5i$

Step-by-step explanation:

i is an imaginary unit, this means

$i^2=-1$

Rewrite -5 as $5\cdot (-1)=5\cdot i^2,$

then

$\sqrt{-5}=\sqrt{5i^2}=\sqrt{5}\cdot \sqrt{i^2}=\sqrt{5}\cdot i=i\sqrt{5}$

Rewrite -25 as $-25=25\cdot (-1)=25\cdot i^2,$

then

$\sqrt{-25}=\sqrt{25\cdot i^2}=\sqrt{25}\cdot \sqrt{i^2}=5\cdot i=5i$

6 0

3 years ago

Prove that the figure is a parallelogram. If so, classify the parallelogram. Be specific and justify your answer.

sesenic [268]

Answer:

Step-by-step explanation:

If it is a parallelogram the opposite sides will a have the same slope.

Using the diagram we see from the coordinates of A and B:

Slope of AB = (5 - -1)/(-1 - -5)

= 6/4

= 3/2.

In the same way

slope of CD = (2 - -4) / (1 - -3)

= 3/2.

So AB and CD can be shown to be parallel.

Similarly the lines BC and AD are parallel.

So the figure is a parallelogram

Finding the perimeter (counting the units between the points):

Perimeter = 2AB + 2BC

By Pythagoras:

AB = sqrt (6^2 + 4^2) = sqrt 52

BC = sqrt (3^2 + 2^2) = sqrt 13

So Perimeter = 2sqrt52 + 2sqrt13

= 4sqrt13 + 2 sqrt13

= 6sqrt13

or 21.63 unit^2 to 2 decimal places.

Area = sqrt52 * perpendicular distance between the lines AB and CD.

8 0

3 years ago

Solve using elimination.<br><br> –9x − 7y = –10<br> 9x + 8y = 5

guapka [62]

Answer:

x=4y

Step-by-step explanation:

mathpapa

7 0

3 years ago

Find the derivative of the following. please show the steps when you answer :1) f(x) = 8xe^x2) y= 5xe^x^43) f(x)= x^8+5/x4) f(t)

borishaifa [10]

Answer:

Since,

$\frac{d}{dx}x^n = nx^{n-1}$

$\frac{d}{dx}(f(x).g(x)) = f(x).\frac{d}{dx}(g(x)) + g(x).\frac{d}{dx}(f(x))$

$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x).f'(x) - f(x) g'(x)}{(g(x))^2}$

1) $y = 8x e^x$

Differentiating with respect to x,

$\frac{dy}{dx}=8( x \times e^x + e^x) = 8(xe^x + e^x) = 8e^x(x+1)$

2) $y = 5x e^{x^4}$

Differentiating w. r. t x,

$\frac{dy}{dx}=5(x\times 4x^3 e^{x^4}+e^{x^4})=5e^{x^4}(4x^4+1)$

3) $y = x^8 + \frac{5}{x^4}$

Differentiating w. r. t. x,

$\frac{dy}{dx}=8x^7 - \frac{5}{x^5}\times 4 = 8x^7 - \frac{20}{x^5}=\frac{8x^{12}-20}{x^5}$

4) $f(t) = te^{11}-6t^5$

Differentiating w. r. t. t,

$f'(t) = e^{11} - 30t^4$

5) $g(p) = p\ln(2p+3)$

Differentiating w. r. t. p,

$g'(p) = p\frac{1}{2p+3}(2) + \ln(2p+3) = \frac{2p}{2p+3}+\ln(2p+3)$

6) $z = (te^{6t}+e^{5t})^7$

Differentiating w. r. t. t,

$\frac{dz}{dt}=7(te^{6t}+e^{5t})^6 ( 6te^{6t}+e^{6t} + 5e^{5t})$

7) $w =\frac{2y + y^2}{7+y}$

Differentiating w. r. t. y,

$\frac{dw}{dy} = \frac{(7+y)(2+2y)-(2y+y^2)}{(7+y)^2} = \frac{14 + 2y + 14y +2y^2 - 2y - y^2}{(7+y)^2}=\frac{14+14y+y^2}{(7+y)^2}$

7 0

4 years ago