What value of k makes the two expressions equivalent: K*x +6 = x +2

Without using a calculator convert the decimal into a fraction 0. 75

mojhsa [17]

Step-by-step explanation:

Decimals can always be put over 1, as they're part of a whole number.

$\frac{0.75}{1}$

Multiply both numerator and denominator by 100:

$\frac{0.75 \times 100}{1\times 100} =\frac{75}{100}$

$\frac{0.75}{1} \: and \: \frac{75}{100} \: are \: fractions \: of \: 0.75$

3 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

3 years ago

Geometry help please

Nana76 [90]

Answer:

CED = 117

Step-by-step explanation:

21x +21 = 7x +49

21x - 7x = 49 - 21

14x = 28

x = 28/14

x = 2

7x + 49

=7(2) + 49

=14 + 49

=63

180 -63 = 117

CED= 117

8 0

2 years ago

Read 2 more answers

8(2f-3)=4f Can anybody give an explanation and answer ? Algebra 1 ?

baherus [9]

Answer:

2

Step-by-step explanation:

16f - 24 = 4f

12f = 24

f = 2

4 0

3 years ago

The distribution of the amount of money spent by first-time gamblers at a major casino in las vegas is approximately normal in s

Lostsunrise [7]

Solution: We are given:

$\mu=600, \sigma =120$

We need to find the z value corresponding to probability 0.84, in order to find the how much money almost 84% of gamblers spent at casino.

Using the standard normal table, we have:

$z(0.85) = 0.9945$

Now we will use the z score formula to find the required amount:

$z=\frac{x-\mu}{\sigma}$

$0.9945=\frac{x-600}{120}$

$0.9945 \times 120 = x - 600$

$119.34 = x - 600$

$x = 600 + 119.34$

$x = 719.34$

$x = 720$ approximately

Therefore, almost 84% of gamblers spent more than $720 amount of money at this casino.

5 0

3 years ago