All categories

2 years ago

I need help for this question please!!!!

Mathematics

1 answer:

Arisa [49]2 years ago

6 0

Answer:

a. 35 degrees

b. 145 degrees

Step-by-step explanation:

a. Since the two base angles of this triangle are congruent, we can conclude that the triangle is isosceles, which means that the two base angles and sides are congruent.

Now, knowing that information, we can subtract 110 from 180 (the sum of all interior angles in a triangle) and we get 70. But this isn't our answer. This is the sum of both base angles. Since the base angles are congruent, we can divide the 70 by 2 to get the measure of ONE base angle, which is 35 degrees.

b. There are two approaches to solve this problem. I have worked both out.

1) We can use the Exterior Angle Theorem, which states that the sum of the interior angles is equal to the exterior angle. We can add 110 to 35, so we get 145 degrees as the measure of <1.

2) The second approach is supplementary angles. Since we see that one of the base angles and <1 is on the same line, we can subtract 35 from 180 to find the measure of <1 to get 145 degrees.

Either way you use, you get the correct answer. Hope this helped!

You might be interested in

A worker can assemble 3 shelf units each hour. During the fifth hour of the shift, there were 115 shelf units in the warehouse.

nasty-shy [4]

A worker can assemble 3 shelf units each hr. During the 5th hr...so the shift has been going on for 5 hrs...at 3 units per hr = (3 * 5) = 15 units assembled during the shift....and if there was 115 units in the warehouse, then that means before the shift, there were (115 - 15) = 100 units assembled <=

4 0

3 years ago

Given the equation 2 square root x-5=2 solve for x and identify if it is an extraneous solution

umka2103 [35]

2√(x-5) = 2
√(x-5) = 1
x-5 = 1
x=6(I don't know what extraneous solution means though)

7 0

2 years ago

Read 2 more answers

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

2 years ago

hi:) I’m so sorry , i uploaded this question earlier but I think the picture couldn’t load. Yea I need help with this question,

love history [14]

Answer:

(ii) $\frac{1}{2} - \frac{1}{3}$

(iii) $\frac{25}{51}$

Step-by-step explanation:

Please see the attached picture for full solution.

7 0

3 years ago

Tan(-x)csc(-x)/sec(-x)cot(-x) simplify the expression

svlad2 [7]

$\bf \cfrac{tan(-x)csc(-x)}{sec(-x)cot(-x)}\implies \cfrac{\quad \frac{sin(-x)}{cos(-x)}\cdot \frac{1}{sin(-x)}\quad }{\frac{1}{cos(-x)}\cdot \frac{cos(-x)}{sin(-x)}}\implies \cfrac{\quad \frac{1}{cos(-x)}\quad }{\frac{1}{sin(-x)}} \\\\\\ \cfrac{1}{cos(-x)}\cdot \cfrac{sin(-x)}{1}\implies \cfrac{sin(-x)}{cos(-x)}\implies \cfrac{\stackrel{\stackrel{symmetry}{identities}}{-sin(x)}}{cos(x)}\implies -tan(x)$

8 0

3 years ago