All categories

3 years ago

5% times blank equals 2.5 NEED ASAP

Mathematics

2 answers:

stealth61 [152]3 years ago

6 0

Answer:

Step-by-step explanation:

5%=0.05

0.05×n=2.5 Is equivalent to 2.5÷0.05=n

n=50

easier way

0.05n=2.5

÷0.05 ÷0.05

n=50

lapo4ka [179]3 years ago

5 0

Answer:

0.5

Step-by-step explanation:

You might be interested in

If I have a mass of 40g and a volume of 20ml, then what is the density?

nydimaria [60]

Answer:

2g/mL^3

Step-by-step explanation:

To find density, we use mass/volume. 40g is the mass and 20mL is the volume. 40/20 is 2, therefore, 2g/mL^3

6 0

3 years ago

Read 2 more answers

“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

Please help ASAP! Please PLEase

Amiraneli [1.4K]

Answer:I'm 80% sure it is c?!?!

Step-by-step explanation:

5 0

2 years ago

Twenty-five percent of what number is 80?

tankabanditka [31]

The answer is 20
Hope this helps

3 0

2 years ago

Read 2 more answers

I don’t know what the last one is

joja [24]

But its still not that much info what is the answers for it if there is no answer

How are you finding what does x equal

8 0

3 years ago

Read 2 more answers