I need help with the unit rate one and every thing else

den301095 [7]

(20*6)+(12*6)+(18*10)=312 500-312=188 colored pencil's left.

4 0

3 years ago

Sebastian is going to deposit $790 in an account that earns 6.8% interest compounded annually his wife Yolanda will deposit $815

marissa [1.9K]

Answer: sorry dont know what to tell you im struggling too

Step-by-step explanation:

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4 0

2 years ago

John mows 6 lawns a week and makes 90 dollars. If he made 270 dollars how many lawns did he mow??? Please help im really struggl

tangare [24]

Answer:

Step-by-step explanation:

90 × 3=270 dollars

6×3= 18 lawns

7 0

3 years ago

Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

Art [367]

Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$