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2 years ago

The product of a number and 11 is 2

Mathematics

1 answer:

rodikova [14]2 years ago

7 0

The answer should come up to n+11=2.

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A rectangular piece of metal is 25 in longer than it is wide. Squares with sides 5 in long are cut from the four corners and the

Licemer1 [7]

Answer:

The original length was 41 inches and the original width was 16 inches

Step-by-step explanation:

Let

x ----> the original length of the piece of metal

y ----> the original width of the piece of metal

we know that

When squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an open box

The dimensions of the box are

$L=(x-10)\ in\\W=(y-10)\ in\\H=5\ in$

The volume of the box is equal to

$V=(x-10)(y-10)5$

$V=930\ in^3$

$930=(x-10)(y-10)5$

simplify

$186=(x-10)(y-10)$ -----> equation A

Remember that

The piece of metal is 25 in longer than it is wide

$x=y+25$ ----> equation B

substitute equation B in equation A

$186=(y+25-10)(y-10)$

solve for y

$186=(y+15)(y-10)\\186=y^2-10y+15y-150\\y^2+5y-336=0$

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=16

see the attached figure

Find the value of x

$x=16+25=41$

therefore

The original length was 41 inches and the original width was 16 inches

3 0

3 years ago

F(r) = 3.14r2. what is f(5)?

rjkz [21]

Answer:

f(5)= 78.5. I hope this helps!!

5 0

3 years ago

What is the distance between (5,-1) and (5,4)?

Alex_Xolod [135]

Answer:

Step-by-step explanation:

The first number in each co-ordinate is the same so -1 plus 4 is 5 units

5 0

3 years ago

Which is greater 5.97 or 5.946????

Ivenika [448]

Answer:

5.97

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Write the equation of a polynomial of degree 3, with zeros 1, 2 and -1 where f(0)=2

drek231 [11]

<u>Answer:</u>

The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is $x^{3}-2 x^{2}-x+2=0$

<u>Solution:</u>

Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.

We have to find the equation of the above polynomial.

We know that, general equation of 3rd degree polynomial is

$F(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+a c) x-a b c=0$

where a, b, c are roots of the polynomial.

Here in our problem, a = 1, b = 2, c = -1.

Substitute the above values in f(x)

$F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0$

$\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}$

So, the equation is $x^{3}-2 x^{2}-x+2=0$

Let us put x = 0 in f(x) to check whether our answer is correct or not.

$\mathrm{F}(0) \rightarrow 0^{3}-2(0)^{2}-0+2=2$

Hence, the equation of the polynomial is $x^{3}-2 x^{2}-x+2=0$

3 0

2 years ago