All categories

3 years ago

Find the areas of each

Mathematics

1 answer:

hjlf3 years ago

6 0

To find the area of each triangle, you just have to multiply the base(bottom) times the height and then divide your answer by 2. Works for all triangles.

You might be interested in

Nick buys 2 shirts and 4 hats for a total of $44.00. Hours f the hats cost $5.00 each,how much does each shirts costs

skad [1K]

Answer:

Each shirt costs 12$

Step-by-step explanation:

Simple steps :)

1. Times 4 by 5 which equals 20

2. Next subtract 20 by 44 which is 24

3. Now take 24 and divide by 2 (Since there are 2 shirts) this is you answer

One shirt costs 12$

Hope this helped have a GOOD day :)

8 0

3 years ago

Plz help me step by step

dybincka [34]

Answer:

x = - 6

Step-by-step explanation:

Equation

1/3 * (-9x +12) = 22 Multiply both sides by 3

Solution

3 * 1/3(-9x + 12) = 22*3

- 9x + 12 = 66 Subtract 12 from both sides.

-9x = 66 - 12 Combine

-9x = 54 Divide by -9

x = 54/-9

x = - 6

6 0

3 years ago

Pleeease open the image and hellllp me

Verdich [7]

1. Rewrite the expression in terms of logarithms:

$y=x^x=e^{\ln x^x}=e^{x\ln x}$

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )

$y'=e^{x\ln x}(x\ln x)'=x^x(x\ln x)'$

$y'=x^x(x'\ln x+x(\ln x)')$

$y'=x^x\left(\ln x+\dfrac xx\right)$

$y'=x^x(\ln x+1)$

2. Chain rule:

$y=\ln(\csc(3x))$

$y'=\dfrac1{\csc(3x)}(\csc(3x))'$

$y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)$

$y'=-3\sin(3x)\cot^2(3x)$

Since $\cot x=\frac{\cos x}{\sin x}$ , we can cancel one factor of sine:

$y'=-3\dfrac{\cos^2(3x)}{\sin(3x)}=-3\cos(3x)\cot(3x)$

3. Chain rule:

$y=e^{e^{\sin x}}$

$y'=e^{e^{\sin x}}\left(e^{\sin x}\right)'$

$y'=e^{e^{\sin x}}e^{\sin x}(\sin x)'$

$y'=e^{e^{\sin x}+\sin x}\cos x$

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:

$\log_2x=\dfrac{\ln x}{\ln2}$

Then

$(\log_2x)'=\left(\dfrac{\ln x}{\ln 2}\right)'=\dfrac1{\ln 2}$

So we have

$y=\cos^2(\log_2x)$

$y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'$

$y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'$

$y'=-\dfrac2{\ln2}\cos(\log_2x)\sin(\log_2x)$

and we can use the double angle identity and logarithm properties to condense this result:

$y'=-\dfrac1{\ln2}\sin(2\log_2x)=-\dfrac1{\ln2}\sin(\log_2x^2)$

5. Differentiate both sides:

$\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'$

$2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\dfrac{xy'}y+\ln y=0$

$-\left(2y-\sin x\,e^y-\dfrac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)$

$y'=\dfrac{2x+\cos x\,e^y\ln y}{2y-\sin x\,e^y-\frac xy}$

$y'=\dfrac{2xy+\cos x\,ye^y\ln y}{2y^2-\sin x\,ye^y-x}$

6. Same as with (5):

$\left(\sin(x^2+\tan y)+e^{x^3\sec y}+2x-y+2\right)'=0'$

$\cos(x^2+\tan y)(x^2+\tan y)'+e^{x^3\sec y}(x^3\sec y)'+2-y'=0$

$\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0$

$\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2\right)$

$y'=-\dfrac{2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2}{\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1}$

7. Looks like

$y=x^2-e^{2x}$

Compute the second derivative:

$y'=2x-2e^{2x}$

$y''=2-4e^{2x}$

Set this equal to 0 and solve for x :

$2-4e^{2x}=0$

$4e^{2x}=2$

$e^{2x}=\dfrac12$

$2x=\ln\dfrac12=-\ln2$

$x=-\dfrac{\ln2}2$

7 0

2 years ago

Each set of three numbers represents the lengths, in units, of the sides of a triangle. Which set can not be used to make a tria

Readme [11.4K]

Answer:

A. 7,6,14

Step-by-step explanation:

Rules for side lengths of triangle.

1. Any side should be less then the sum of the other two angles.

2. The same side should be greater than the subtraction of the other side lengths(bigger side - smaller side)

7 < 6+14 . Correct

7> 14-6 . Incorrect

This cannot be a solution (impossible)

4<4+4 . Correct

4>4-4 . correct

This is a solution (equilateral triangle)

6<6+2 . Correct

6>6-2 . Correct

This is a solution (isosceles triangle)

7<8+13 . Correct

7>13-8 . Correct

This is a solution (scalene triangle)

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

3 0

2 years ago

Two square pyramids are similar. If the ratio of a pair of corresponding edges is a : b, what is the ratio of their volumes? Wha

BabaBlast [244]

Ratio of their volumes = a³ : b³

Ratio of their surface areas = $$\frac{a^{2} + a\sqrt{4h^{2}+a^{2}}}{b^{2} + b\sqrt{4h^{2}+b^{2}}}$

Step-by-step explanation:

Two square pyramids are similar with their edges are in the ratio of a : b.

Volume of a square pyramid with edge and h = a is given by the formula,

= $$\frac{a^{2}\times h }{3}$ = $$\frac{a^{3} }{3}$

Volume of a square pyramid with edge b and h = b is given by the formula,

= $$\frac{b^{2}\times h }{3}$ = $$\frac{b^{3} }{3}$

Ratio of their volumes = a³ : b³ since h/3 gets cancelled.

Total surface area of square pyramid with the edge a =

$$a^{2} + a\sqrt{4h^{2}+a^{2}}$

Total surface area of square pyramid with the edge b =

$$b^{2} + b\sqrt{4h^{2}+b^{2}}$

Ratio of the surface area = $$\frac{a^{2} + a\sqrt{4h^{2}+a^{2}}}{b^{2} + b\sqrt{4h^{2}+b^{2}}}$

8 0

3 years ago