All categories

2 years ago

One angle of a triangle measures 65°. The other two angles are in a ratio of 3:20. What are the measures of those two angles?

Mathematics

1 answer:

Nataly_w [17]2 years ago

3 0

Answer:

15° and 100°

Step-by-step explanation:

$65 + 3x + 20x = 180$

$23x = 115$

$x = 5$

3 × 5 = 15, 20 × 5 = 100. So the other two angles measure 15° and 100°.

You might be interested in

467.4 + 0.492 =<br><br>PLS HELP

ratelena [41]

Answer:

467.892

Step-by-step explanation:

467.4+0.492 =467.892

3 0

3 years ago

Read 2 more answers

A square garden plot measures 180 square feet. A second square garden plot measures 320 square feet. How many more feet of fence

REY [17]

The second garden plot will require 8√5 feet more fence than the first garden plot.

Further explanation:

In order to find the fence, we have to find the perimeter of both squares

So,

Area of Square 1: A1=180 square feet

Area of Square 2: A2=320 Square feet

Let x be the side of square 1:

Then,

$A_1=x^2\\180=x^2\\Taking\ square\ root\ on\ both\ sides\\\sqrt{180}=\sqrt{x^2}\\x=\sqrt{2*2*3*3*5}\\x=\sqrt{2^2*3^2*5}\\x=2*3\sqrt{5}\\x=6\sqrt{5}$

For second square:

Let y be the side of second square

$A_2=y^2\\320=y^2\\Taking\ square\ root\ on\ both\ sides\\\sqrt{320}=\sqrt{y^2}\\y=\sqrt{2*2*2*2*2*2*5}\\y=\sqrt{2^2*2^2*2^2*5}\\y=2*2*2\sqrt{5}\\y=8\sqrt{5}$

Perimeter of First Square:

$P_1=4x\\=4(6\sqrt{5})\\=24\sqrt{5}\ feet$

Perimeter of Second Square:

$P_2=4y\\=4(8\sqrt{5})\\=32\sqrt{5}\ feet$

The smaller perimeter will be subtracted from larger perimeter to find that how much more fence will be needed.

$P_2-P_1=32\sqrt{5}-24\sqrt{5}\\=(32-24)\sqrt{5}\\=8\sqrt{5}\ feet$

The second garden plot will require 8√5 feet more fence than the first garden plot.

Keywords: Radicals, Operations on Radicals

Learn more about radicals at:

#LearnwithBrainly

6 0

4 years ago

In many fast food restaurants, there is a strong correlation between a menu item's fat content (measured in grams) and its calor

Kruka [31]

Answer:

1. The slope of the least-squares regressin line is 11.364

2. The intercept of the least-squares regression line is 204.343

3. The residual corresponding to this observation is -10 calories

Step-by-step explanation:

1. In order to calculate the slope of the least-squares regressin line we would have to use the following formula:

slope of the least-squares regressin line=r*sy/sx

Therefore, slope=0.979*324.90/27.99

slope=11.364

2. In order to calculate the intercept of the least-squares regression line we would have to use the following formula:

intercept of the least-squares regression line=ybar - slope*xbar

intercept=662.88 - 11.364*40.35

intercept=204.343

3. In order to calculate the the residual corresponding to this observation we would have to use the following formula:

residual = observed y - predicted y

predicted y=11.364*x + 204.343

if the x value is 7, therefore predicted y=11.364*107 + 204.343

predicted y=1,420

Therefore, residual=1410 calories - 1420 calories

residual=-10 calories

8 0

4 years ago

What are the solutions of 2x^2 + 16x + 34 = 0?

nexus9112 [7]

Answer :

2x^2 + 16x + 34 = 0? X= -4 -1i X= 4+1i

Step-by-step explanation:

6 0

3 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

4 years ago