All categories

3 years ago

What is the third angle in a triangle with two angles of 60* and 45*

Mathematics

1 answer:

REY [17]3 years ago

6 0

The answer is 75° because all three angles of a triangle must add to 180°

You might be interested in

Hexagon DEFGHI is translated on the coordinate plane below to create hexagon D'E'F'G'H'I': Hexagon DEFGHI and Hexagon D prime E

MatroZZZ [7]

<span>(x, y)→(x − 8, y − 7) is the correct translation. Take point D for example. The coordinate of D is (2, 5). The coordinate of D' (after translation) is (-6, -2). Since 2-8=-6, 5-7=-2, only the first choice is correct. You can also try other points and see why only this is the right translation.</span>

4 0

3 years ago

Read 2 more answers

Evaluate −3w − 6p for w=2 and p = −7

kap26 [50]

-3w-6p when w=2 and p=-7

-3(2)-6(-7)

= -6 + 42

= 36

7 0

2 years ago

Read 2 more answers

Find the measure of each angle indicated. Round to the nearest tenth

MArishka [77]

Answer:

1. A 2.C

Step-by-step explanation:

8 0

2 years ago

(2x+4)^2 - (x-5)^2 = 26x<br> Please solve (explained)

Zina [86]

Expand the squared binomials:

$(2x+4)^2=4x^2+16x+16$

$(x-5)^2=x^2-10x+25$

Then

$(4x^2+16x+16)-(x^2-10x+25)=3x^2+26x-9=26x$

$\implies3x^2-9=0$

$\implies3x^2=9$

$\implies x^2=3$

$\implies x=\pm\sqrt3$

5 0

3 years ago

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$

$A = \frac{L}{k} - \frac{L}{k} e^{kt}$

So, D = R - A =

$D = \frac{L}{k} - \frac{L}{k} - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}$

when t = 0(at initial time), the initial value of D =

$D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}$

4 0

3 years ago