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2 years ago

13

Factor the expression r^2+19rs+90s^2

2 answers:

Natasha2012 [34]2 years ago

8 0

The expression factorises to C.

daser333 [38]2 years ago

3 0

The answer to the question is c

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Two ladders AB and PQ are resting against opposite

muminat

Answer:

The length of the longer ladder is 35 ft Step-by-step explanation: Please check the attachment for a diagrammatic representation of the problem We want to calculate the length of the longer ladder ; We make reference to the diagram Since the two right triangles formed are similar. the ratios of their sides are equal; Thus; 20/15 = 28/x + 15 20(x + 15) = 15(28) 20x + 300 = 420 20x = 420-300 20x = 120 x = 120/20 x = 6 So we want to calculate the hypotenuse of a right triangle with other sides 28ft and 21 ft To do this, we use the Pythagoras’ theorem which states that square of the hypotenuse equals the sum of the squares of the two other sides Let the hypotenuse be marked x x^2 = 28^2 + 21^2 x^2 = 1,225

x = √1225

x = 35 ft

5 0

1 year ago

Please help I’m about to fail and this quiz will bring my grade up.

konstantin123 [22]

Answer:

first blank: -1

second blank: -7

third blank: -13

fourth blank: - 19

Step-by-step explanation:

hope this helps and please rate my answer!

5 0

2 years ago

2 thousands divided by what number is equal to 2 hundreds?

laila [671]

Answer:

10

Step-by-step explanation:

it is 10 2000÷10=200

8 0

2 years ago

Barrett works at an ice cream shop. The function f(x) represents the amount of money in dollars Barrett earns per gallon of ice

Paha777 [63]

Answer:

$f(g(x))=6x^3} +4$

Step-by-step explanation:

Given:

$f(x)=2x^2+4\\\\g(x)=\sqrt{3x^3}$

Required:

f(g(x))=?

Solution:

let f(g(x))=f(X), where X=g(x)

so $f(X)=2X^2+4$

put X= $g(x)=\sqrt{3x^3}$ , we get

$f(g(x))=2(\sqrt{3x^3} )^2+4$

$f(g(x))=2(\sqrt{3x^3} )^2+4\\\\f(g(x))=2({3x^3} )+4\\\\f(g(x))=6x^3} +4$

5 0

2 years ago

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

2 years ago