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a triangular sail has sides of 10 ft ,24ft,and 26ft .if the shortest side of a similar sail measures 6ft , what is the measure o

Katarina [22]

Answer:

15.6 ft

Step-by-step explanation:

Since the sails are similar then the ratios of corresponding sides are equal.

Comparing the shortest sides, that is

10 : 6 = 5 : 3

let the longest side be x, then using proportion

$\frac{26}{5}$ = $\frac{x}{3}$ ( cross- multiply )

5x = 78 ( divide both sides by 5 )

x = 15.6

The longest side is 15.6 ft

5 0

3 years ago

Two factors whose product is 84 are 12 and _[blank]_.

djyliett [7]

Answer: 7

Step-by-step explanation:

To solve this problem you must apply the proccedure shown below:

1. You know that a multiplication has the following form:

a*b=c

Where a and b are the factors and c is the product.

2. You know one of the factor, then you can find the second one as following:

12*b=84

b=84/12

b=7

Therefore the answer is 7.

7 0

4 years ago

-2 (3 + 5y -10) = 34

IrinaK [193]

-2 (3+5y-10) = 34
Step 1) -6-10y+20=34
Step 2) -10y+20=40
Step 3) -10y=20
Step 4) y = -2
Answer is -2

7 0

3 years ago

Read 2 more answers

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

4 years ago

(6z 2 - 4z + 1)(8 - 3z).

Yakvenalex [24]

Distributive property
a(b+c)+ab+ac
a(b-c)=ab-ac

(6z^2-4z+1)(8-3z)
move for nicety
(8-3z)(6z^2-4z+1)
distribute
8(6z^2-4z+1)-3z(6z^2-4z+1)=
48z^2-32z+8-18z^3+12z^2-3z=
-18z^3+48z^2+12z^2-32z-3z+8=
-18z^3+60z-35z+8

7 0

3 years ago