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2 years ago

Which of the follwing is correxr

Mathematics

1 answer:

agasfer [191]2 years ago

3 0

Answer:

the equation has infinitely amny solutions

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NEED URGENT HELP!

sp2606 [1]

(3,9)(5,15)
slope = (15 - 9) / (5 - 3) = 6/2 = 3

y = mx + b
slope(m) = 3
(3,9)...x = 3 and y = 9
sub and find b, the y int
9 = 3(3) + b
9 = 9 + b
9 - 9 = b
0 = b

so ur equation is : y = 3x + 0....can be written as y = 3x.

after 12 months...
y = 3x...x = 12
y = 3(12)
y = 36 inches

6 0

3 years ago

Let f(x) = x^3 + 2x^2 + 3x + 1 and g(x)= 4x-5 Find f(x) x g(x)

Brums [2.3K]

$▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ { \huge \mathfrak{Answer}}▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪$

Given :

$f(x) = {x}^{3} + 2 {x}^{2} + 3x + 1$

$g(x) = 4x - 5$

let's solve for f(x) × g(x) :

$( {x}^{3} + 2x {}^{2} + 3x + 1)(4x - 5)$

$4x {}^{4} - 5 {x}^{3} + 8 {x}^{3 } - 10 {x}^{2} + 12 {x}^{2} - 15x + 4x - 5$

$4 {x}^{4} + 3 {x}^{3} + 2 {x}^{2} - 11x - 5$

therefore, correct choice is : D

8 0

2 years ago

Evaluate (2.0 x 10^-4) x (3.0x10^-3

Mars2501 [29]

Answer:

6*10^-7

Step-by-step explanation:

5 0

2 years ago

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor

Artist 52 [7]

Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

$C=2\pi r$

Find the radius of the sphere with a 5.50-meter circumference

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

$r=5.50/[2(3.14)]=0.88\ m$

Find the radius of the sphere with a 7.85-meter circumference

For $C=7.85\ m$

assume

$\pi =3.14$

substitute and solve for r

$7.85=2(3.14)r$

$r=7.85/[2(3.14)]=1.25\ m$

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

$SA=4\pi r^{2}$

Find the surface area of sphere with a 5.50-meter circumference

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(0.88)^{2}$

$SA=9.73\ m^{2}$

Find the surface area of sphere with a 7.85-meter circumference

For $r=1.25\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(1.25)^{2}$

$SA=19.63\ m^{2}$

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

$9.73*(92)=\$895.16$

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

$19.63*(92)=\$1,805.96$

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

6 0

2 years ago

253,761 In Expanded Form .

Irina18 [472]

2 hundred thousands
5 ten thousands
3 thousands
7 hundreds
6 tens
1 ones

6 0

3 years ago