A storage container shaped as a rectangular prism can hold 896 wooden cube blocks with edge lengths of 14 ft.

The larger of two numbers is 3 more than twice the smaller. if the twice the larger is decreased by 5 times the smaller the resu

Natasha2012 [34]

Let x be the large number, and y be the smaller one: x=3+2y, 2x-5y=0 (from the question) --> substitute the first equation in the second --> 2(3+2y)-5y=0 --> 6+4y-5y=0 -->6-y=0 --> y=6, substitute y=6 in the first equation --> x=3+2(6) --> x=15. therefore the two numbers are 6,15

4 0

3 years ago

The value of zlatans house has increased by 7% His house is now valued at ?749,000 Work out the value of the house before the in

wolverine [178]

Answer:

$440588.24

Step-by-step explanation:

Let x be the value of Zlatans house before the increase.

We are told that the value of Zlatans house has increased by 7%. His house is now valued at $749,000. This means that value of house(x) before increase plus 7% of x will be equal to $749000.

We can set this information in an equation as:

$x+\frac{70}{100}x=74900$

$x+0.7x=74900$

Upon combining like terms we will get,

$(1+0.7)x=74900$

$1.7x=74900$

$x=\frac{749000}{1.7}$

$x=440588.2352941176470588\approx 440588.24$

Therefore, the value of the house before increase will be $440588.24.

4 0

3 years ago

In T-ball, the distance to each successive base is 50 feet. If the distance from home plate to the pitcher’s mound is 38 feet, h

JulijaS [17]

Answer:

<h2>The distance from the pitcher's mound and to second base is 37.99 approximately.</h2>

Step-by-step explanation:

The diamond is a square, which in this case has 50 feet long each side, and from home to pitcher is 38 feet. Notice that home is a vertex of the square and the pitcher's mound is the intersection of the diagonals, where they cut half.

We can find the distance from the pitcher to first base using Pythagorean's Theorem, where 50 feet is the hypothenuse.

$50^{2} =38^{2}+x^{2}\\x^{2}=50^{2}-38^{2}\\x=\sqrt{2500-1444}\\ x=\sqrt{1056}\\ x \approx 32.5 \ ft$

Therefore, the distance from the pitcher to first base is 32.5 feet, approximately.

Now, we can use again Pythagorean's Theorem to find the distance from pitcher to second base, where the hypothenuse is 50 feet.

$50^{2}=32.5^{2}+y^{2}\\y^{2}=50^{2}-32.5^{2}\\y=\sqrt{2500-1056.25}\\ y =\sqrt{1443} \approx 37.99$

Therefore, the distance from the pitcher's mound and to second base is 37.99 approximately.

<em>(this results make sense, because the diagonals of a square intersect at half, that means all bases have the same distance from pitcher's mound, so the second way to find the distance asked in the question is just using theory)</em>

8 0

3 years ago

The newly elected president needs to decide the remaining 8 spots available in the cabinet he/she is appointing. If there are 14

Elanso [62]

Answer:

The members of the cabinet can be appointed in 121,080,960 different ways.

Step-by-step explanation:

The rank is important(matters), which means that the order in which the candidates are chosen is important. That is, if we exchange the position of two candidates, it is a new outcome. So we use the permutations formula to solve this quesiton.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

If there are 14 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed?

Permutations of 8 from a set of 14. So

$P_{(14,8)} = \frac{14!}{(14-8)!} = 121,080,960$

The members of the cabinet can be appointed in 121,080,960 different ways.

3 0

3 years ago

Students at Westbrook middle school hold green team meetings

Tems11 [23]

What’s the whole question???!

7 0

3 years ago