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1 year ago

One leg of a 90-45-45 triangle is 5. What are the sides of the hypotenuse and the other leg?

Mathematics

1 answer:

kirza4 [7]1 year ago

7 0

Answer:

5√2, 5

Step-by-step explanation:

in a 90 45 45 triangle, each leg is the same legnth

the hypotenuse of a 90 45 45 triangle is x√2

x being the measure of the leg

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Is this a function? {(-1,3); (-2,-3); (2,0); (0,2)}

Molodets [167]

Yes, because the x’s aren’t repeated

8 0

2 years ago

Determine if the two triangles are congruent if they are state how do you know SSS,SAS,HL,ASA, OR AAS. Be sure to include any ex

dlinn [17]

1: yes AAS
2: yes SSS reflective
3: yes ASA
4: yes HL reflective
5: no SSA vertical
6: yes SAS vertical

Not sure on vocabulary for 1 and 3 sorry:(

7 0

2 years ago

The diameter of a circle is 1.4 centimeters. Find its area. Use 3.14 Round to the nearest

NeTakaya

Answer:

Area = 1.54

Step-by-step explanation:

Area = π $r^{2}$

π = 3.14

diameter = 1.4

radius (half of diameter) = 0.7

Area = 3.14 x $0.7^{2}$

Area = 3.14 x 0.49

Area = 1.5386

round to the nearest hundredth:

1.5386 = 1.54

8 0

2 years ago

Read 2 more answers

Kyle works at a donut factory, where a 10-oz cup of coffee costs 95¢, a 14-oz cup costs $1.15, and a 20-oz cup costs $1.5

Fynjy0 [20]

Answer:

Kyle filled 4 10-oz cups, 6 14-oz cups, and 4 20-oz cups.

Step-by-step explanation:

Let 10-oz, 14-oz, and 20-oz coffees be represented by the variables a, b, and c, respectively.

Since a total of 14 cups of coffee was served:

$a+b+c=14$

A total of 204 ounces of coffee was served. Therefore:

$10a+14b+20c=204$

A total of $16.70 was collected. Hence:

$0.95a+1.15b+1.5c=16.7$

This yields a triple system of equations. In order to solve a triple system, we should isolate the system to only two variables first.

From the first equation, let's subtract a and b from both sides:

$c=14-a-b$

Substitute this into both the second and third equations:

$10a+14b+20(14-a-b)=204$

And:

$0.95a+1.15b+1.5(14-a-b)=16.7$

In this way, we've successfully created a system of two equations, which can be more easily solved. Distribute:

For the Second Equation:

$\displaystyle \begin{aligned} 10a+14b+280-20a-20b&=204\\ -10a-6b&=-76\\5a+3b&=38\end{aligned}$

And for the Third:

$\displaystyle \begin{aligned} 0.95a+1.15b+21-1.5a-1.5b&=16.7\\ -0.55a-0.35b&=-4.3\end{aligned}$

We can solve this using substitution. From the second equation, isolate a:

$\displaystyle a=\frac{1}{5}(38-3b)=7.6-0.6b$

Substitute into the third:

$-0.55(7.6-0.6b)-0.35b=-4.3$

Distribute and simplify:

$-4.18+0.33b-0.35b=-4.3$

Therefore:

$-0.02b=-0.12\Rightarrow b=6$

Using the equation for a:

$a=7.6-0.6(6)=4$

And using the equation for c:

$c=14-(4)-(6)=14-10=4$

Therefore, Kyle filled 4 10-oz cups, 6 14-oz cups, and 4 20-oz cups.

7 0

2 years ago

4.20. According to a report released by the National Center for Health Statistics, 51% of U.S. households has only cell phones (

Sergeu [11.5K]

Answer:

The probability that the household has only cell phones and has high-speed Internet is 0.408

Step-by-step explanation:

Let A be the event that represents U.S. households has only cell phones

Let B be the event that represents U.S. households have high-speed Internet.

We are given that 51% of U.S. households has only cell phones

P(A)=0.51

We are given that 70% of the U.S. households have high-speed Internet.

P(B)=0.7

We are given that U.S. households having only cell phones, 80% have high-speed Internet. A U.S household is randomly selected.

P(B|A)=0.8

$\frac{P(A\capB)}{P(A)}=0.8\\P(A\capB)=0.8 \times P(A)\\P(A\capB)=0.8 \times 0.51\\P(A\capB)=0.408$

Hence the probability that the household has only cell phones and has high-speed Internet is 0.408

7 0

2 years ago