MATHEMATICS HIGH SCHOOL

The Martin family is considering building a new house. On the blueprints, the dimensions of the outside of the house are 18 inches wide by 26 inches long. If the house is actually going to be 45 feet wide, what will the actual length of the house be?

Answers

Answer 1

Answer:

This can be answered through simple ratio and proportion

Let x = width

y = height

x/y = x2/y2 = x3/y3

since

for 18 inches width we have 26 inches as length, therefore

18/26 = 45/y2

y2 = 65 ft

there is no need to convert the units for the smaller dimensions since we are dealing with definite ratios thus unitless.

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I really need help, I hope you can help me out please

Answers

Answer:

Step-by-step explanation:

they are not similar -

the triangles angles are not congruent which then means that by AA criterion the triangles are not similar.

hope this helps!!

Answer:

Yes they are similar as the obtuse angle of 120 degree is the same and 2 other angles are acute and this example shows 4 degree range in difference. This would change the measure 0.4 only each side.

Similar is different to same, the exact meaning of similar is resemblance, the exact meaning is having a resemblance in appearance, character, or quantity, without being identical.

Step-by-step explanation:

Jill is building a wooden raised bed garden for her vegetable she wants the frame to have dimension X feet by X +3 feet if she wants the garden to have a maximum area of 50 ft.² what are the possible values of X

Answers

Answer: 0ft < X ≤ 5.73ft

Step-by-step explanation:

In this case, we have a rectangle.

For a rectangle of length L and width W, the area is:

A = L*W.

And we have:

L = X

W = X + 3ft.

Then the area will be:

A = X*(X + 3ft) = X^2 + 3ft*X.

And we want to have a maximum area of 50ft^2.

Then we can write:

A = X^2 + 3ft*X ≤ 50ft^2

Now let's solve this for X.

Now, the first thing we can see is that both coefficients in our quadratic equation are positive, so as the absolute value of X increases, also does the whole equation.

Then makes sense start for the upper limit of X, this is when:

X^2 + 3ft*X = 50ft^2.

Now we can solve the quadratic equation:

X^2 + 3ft*X - 50ft^2 = 0

Applying the Bhaskara formula, the solutions are:

$X = (-3ft +- √((3ft)^2 - 4*1*(-50ft^2)) )/(2) = (-3ft +-14.46ft )/(2)$

Then we have two solutions:

X = (-3ft - 14.46ft)/2 = -8.73 ft.

X = (-3ft + 14.46ft)/2 = 5.73 ft

Because X represents a distance, it can only be positive, then we must select the option X = 5.73ft.

This is the maximum value of X, and we will have:

0ft < X ≤ 5.73ft

Where the lower limit is there because we can not have X = 0ft, as this does not have physical meaning.

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CAN'T YOU HELP ME THE BEAUTIFUL GIRL???!!!

Answers

Answer:

$(1)/(2)$

Step-by-step explanation:

Given: Kristen made 4 out of 5 throws in her first basketball game, 5 out of 8 in her second game and 5 out of 7 free throws in her third game.

To Find: The experimental probability of Kristen marking her first free throw in her fourth game.

Solution:

There are only two possibilities that the first throw will be free or not.

Hence, the total outcomes = 2

Hence, Probability that the throw is free = $(1)/(2)$

Hence, 1/2 is the experimental probability of Kristen marking her first free throw in her fourth game.

If cos θ -sin θ =1, find θ cos θ +sin θ?

Answers

$cos\theta-sin\theta=1\ \ \ \ |square\ both\ sides\n\n(cos\theta-sin\theta)^2=1\n\ncos^2\theta-2sin\theta cos\theta+sin^2\theta=1\n\n1-2sin\theta cos\theta=1\n\n-2sin\theta cos\theta=1-1\n\n-2sin\theta cos\theta=0\iff sin\theta=0\ or\ cos\theta=0\n\n\theta=k\pi\ or\ \theta=(\pi)/(2)+k\pi\ \ \ \ (k\in\mathbb{Z})$

$For\ \theta=k\pi\to cosk\pi-sink\pi=\pm1-0=\pm1\to conclusion:\theta=2k\pi\n\nFor\ \theta=(\pi)/(2)+k\pi\to cos((\pi)/(2)+k\pi)-sin((\pi)/(2)+k\pi)=0-(\pm1)=\mp1\to\n\to conclusion:\theta=(3\pi)/(2)+2k\pi\n\n============================\n\ncos\theta+sin\theta=(*)\n\nfor\ \theta=2k\pi\to(*)=1+0=1\n\nfor\ \theta=(3\pi)/(2)+2k\pi\to(*)=0-1=-1$

The length of a swimming pool is 8m longer than its width and the area is 105m2

Answers

$width : w \n length : \ l = w + 8 \n A = 105 \ m^2 \n \nA= w \cdot l \n \nw(w+l)=105 \n \nw^2+w = 105 \n \nw^2+w-105 =0 \n \n(w+15)(w-7)=0 \n \nw+15 =0 \ \ or \ \ w-7 = 0 \n \nw = -15 \ \ or \ \ w=7 \n \n width \ cant \ be \ negative, \ so \n \n w = 7 \ and \ l = w+8 = 7 +8 =15 \n \nchek : \n \n A=w\cdot l \n \nA=7\cdot 15=105 \ m^2$

$l\ \rightarrow\ the\ length\nw\ \rightarrow\ the\ width \n\nl=w+8\ \ \ and\ \ \ l\cdot w=105\ \ \ and\ \ \ l>0,\ w>0\n\n(w+8)\cdot w=105\ \ \ \Rightarrow\ \ \ w^2+8w=105\n\n w^2+2\cdot4w+4^2=105+4^2\n\n (w+4)^2=105+16\ \ \ \Rightarrow\ \ \ (w+4)^2=121\n\nw+4=11\ \ \ or\ \ \ w+4=-11\n\nw=7\ \ \ \ \ \ \ \ \ \ or\ \ \ w=-15<0\n\nw=7\ \ \ \Rightarrow\ \ \ l=7+8=15\n\nAns.\ the\ length\ is\ 15\ m\ \ and\ \ the\ width\ is\ 7\ m.$

A scale drawing of an office space uses a scale of 2 inches in 3 yards the scale drawing has an area of 16 in.² what is the area of the actual office space