All categories

2 years ago

Will give brainliest a. 39cm^2 b. 169cm^2 c. 338cm^2 d. 676cm^2

Mathematics

2 answers:

zaharov [31]2 years ago

8 0

Answer:

13×36

Step-by-step explanation:

the angle wants you to multiply 13×36

Damm [24]2 years ago

3 0

Answer:

A = 1/2bh

b = 26

h = 13

1/2 * 13*26

338*1/2

169

You might be interested in

Jane has 312 feet of fencing to use to enclose a rectangular garden. She wants the garden to be 5 times as long as it is wide. W

VashaNatasha [74]

Length is 130 ft. and width is 26 ft.

3 0

2 years ago

Find the exact value of cos theta, given that sin thetaequalsStartFraction 15 Over 17 EndFraction and theta is in quadrant II.

vova2212 [387]

Answer:

$cos \theta = -\frac{8}{17}$

Step-by-step explanation:

For this case we know that:

$sin \theta = \frac{15}{17}$

And we want to find the value for $cos \theta$ , so then we can use the following basic identity:

$cos^2 \theta + sin^2 \theta =1$

And if we solve for $cos \theta$ we got:

$cos^2 \theta = 1- sin^2 \theta$

$cos \theta =\pm \sqrt{1-sin^2 \theta}$

And if we replace the value given we got:

$cos \theta =\pm \sqrt{1- (\frac{15}{17})^2}=\sqrt{\frac{64}{289}}=\frac{\sqrt{64}}{\sqrt{289}}=\frac{8}{17}$

For our case we know that the angle is on the II quadrant, and on this quadrant we know that the sine is positive but the cosine is negative so then the correct answer for this case would be:

$cos \theta = -\frac{8}{17}$

5 0

2 years ago

Differentiate the function. f(x) = ln(25 sin^2(x))

tigry1 [53]

Use the chain rule.
Let u = 25sin²(x), such that dy/dx = dy/du · du/dx

$\frac{dy}{dx} = \frac{1}{25sin^{2}(x)} \cdot 50cos(x) \cdot sin(x)$
$\frac{dy}{dx} = \frac{2cos(x) sin(x)}{sin^{2}(x)}$

$\frac{dy}{dx} = 2cot(x)$

8 0

3 years ago

Can someone PLEASE help me solve this equation ? due soon

RideAnS [48]

$\sf{Given : 3tanx + 7 = \dfrac{2}{(1 - sinx)(1 + sinx)}}$

We know that : (a - b)(a + b) = a² - b²

$\implies \sf{3tanx + 7 = \dfrac{2}{1 - sin^2x}}$

We know that : 1 - sin²x = cos²x

$\implies \sf{3tanx + 7 = \dfrac{2}{cos^2x}}$

$\sf{\bigstar \ \ We \ know \ that : \boxed{\sf{\dfrac{1}{cos^2x} = sec^2x}}}$

$\implies \sf{3tanx + 7 = 2sec^2x}$

We know that : sec²x = 1 + tan²x

$\implies \sf{3tanx + 7 =2(1 + tan^2x)}$

$\implies \sf{2 + 2tan^2x - 3 tanx - 7 = 0}$

$\implies \sf{2tan^2x - 3 tanx - 5 = 0}$

$\implies \sf{2tan^2x - 5tanx + 2tanx - 5 = 0}$

$\implies \sf{2tanx(tanx + 1) - 5(tanx + 1) = 0}$

$\implies \sf{(tanx + 1)(2tanx - 5) = 0}$

$\implies \sf{tanx = -1 \ (or) \ tanx = \dfrac{5}{2} }$

8 0

2 years ago

PLEASE HELP, THIS IS REALLY IMPORTANT. PLEASE ANSWER ASAP.

Assoli18 [71]

Yes the diagonals of a parallelogram have the same midpoint since they ... of the intersection of the diagonals of parallelogram AB CD given the vertex points ... If a parallelogram is a rhombus then its diagonals are? , statement 2 is the answer

8 0

2 years ago