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[will mark brainleist]

balandron [24]

Answer:

164 inches

Step-by-step explanation:

A=2(wl+hl+hw)

you subsitute your l width and height

2(3(10)+4(10)+4(3))

2(30+40+12)

2(82) gets you to 164 inches squared

8 0

3 years ago

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4x + y = 3 -x + 2y = -3 I need to use substitution to solve this. help?

IrinaK [193]

Answer:

(x,y) (1,-1)

Step-by-step explanation:

y=3-4x

-x+2y=-3

x+2(3-4x)=-3

x=1

y=3-4*1

y=-1

4 0

3 years ago

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A 41 41-foot ladder is leaning against a building. The base of the ladder is 9 9 feet from the side of the building. How high do

lyudmila [28]

Answer:

The ladder would reach 40ft

Step-by-step explanation:

The ladder forms a right angle triangle with the building and the ground. The length of the ladder represents the hypotenuse of the right angle triangle. The height that the ladder would reach along the side of the building represents the opposite side of the right angle triangle.

The distance from the bottom of the ladder to the base of the building represents the adjacent side of the right angle triangle.

To determine the height that the ladder would reach along the side of the building h, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

41² = 9² + h²

1681 = 81 + h²

h² = 1681 - 81 = 1600

h = √1600

h = 40 ft

6 0

3 years ago

Mr. Thaxton has two daughters. The sum of their ages is twenty-one. The product of their ages is one hundred ten. How old is Mr.

tatyana61 [14]

Answer:

The youngest daughter is 10

6 0

3 years ago

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If you are dealt five cards from a shuffled deck of 52 cards find the probability of getting three queens and two kings

dexar [7]

The probability of getting three queens and two kings is $\frac{1}{1082900}$

Solution:

Given that , you are dealt five cards from a shuffled deck of 52 cards

We have to find the probability of getting three queens and two kings

Now, we know that, in a deck of 52 cards, we will have 4 queens and 4 kings.

$\text { probability of an event }=\frac{\text { favarable possibilities }}{\text { number of possibilities }}$

Probability of first queen:

$\text { Probability for } 1^{\text {st }} \text { queen }=\frac{4}{52}=\frac{1}{13}$

Probability of second queen:

$\text { Plobability for } 2^{\text {nd }} \text { queen }=\frac{3}{51}=\frac{1}{17}$

Here we used 3 for favourable outcome, since we already drew 1 queen out of 4

And now number of outcomes = 52 – 1 = 51

Probability of third queen:

Similarly here favorable outcome = 2, since we already drew 2 queen out of 4

And now number of outcomes = 51 – 1 = 50

$\text { Probability of } 3^{\text {rd }} \text { queen }=\frac{2}{50}=\frac{1}{25}$

Probability for first king:

Here kings are 4, but overall cards are 49 as 3 queens are drawn

$\text { probability for } 1^{\text {st }} \text { king }=\frac{4}{49}$

Probability for second king:

Here, kings are 3 and overall cards are 48 as 3 queens and 1 king are drawn

$\text { probability of } 2^{\text {nd }} \text { king }=\frac{3}{48}=\frac{1}{16}$

And, finally the overall probability to get 3 queens and 2 kings is:

$=\frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} \times \frac{4}{49} \times \frac{1}{16}=\frac{4}{4331600}=\frac{1}{1082900}$

Hence, the probability is $\frac{1}{1082900}$

7 0

3 years ago