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1 year ago

4

Classify the lunar phase that has the near side of the Moon completely illuminated. New Moon First Quarter Moon Full Moon Last Q

uarter Moon

1 answer:

yulyashka [42]1 year ago

8 0

If the near side is fully illuminated, it would be a full moon.

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A butcher is mixing ground turkey breast that is 98% lean and beef that is 90% lean to make a blend that is 92.4% lean. Determin

Paul [167]

See brainly.com/question/8613376 for one way to work mixture problems.

The proportion of beef will be
.. (98 -92.4)/(98 -90) = 5.6/8 = 0.7

70 pounds of beef and 30 pounds of turkey are needed.

_____
1. I would prefer to use "t" and "b" as variables, rather than x and y. That way I can more easilly keep straight that they represent pounds of turkey and pounds of beef, respectively.

2. t + b = 100 . . . . . we want 100 pounds of mix

3. 98%*t +90%*b = 92.4%*100 . . . . . total pounds of lean meat

4. Using subsitution for t, we have
.. 98%*(100 -b) +90%*b = 92.4%*100
.. b*(98 -90) -98*100 = -92.4*100 . . . . . . multiply by -100, collect b terms
.. b = (98 -92.4)*100/(98 -90) . . . . . . . . . add 98*100 and divide by b coefficient
note that this expression is exactly the one we used to write down the answer above.
.. b = 5.6*100/8 = 70 . . . . . . . y, if you like
.. t = 100 -b = 30 . . . . . . . . . . x, if you like

70 pounds of beef and 30 pounds of turkey are needed.

7 0

1 year ago

Lines s, t, and u are perpendicular bisectors of the sides of APGH and meet at J. If JG - 2x+ 2, JH - 2y-4, JF-8 and

Y_Kistochka [10]

Answer:

x = 3, y = 6, z = 4

Step-by-step explanation:

If lines s, t, and u are perpendicular bisectors of the sides of APGH and meet at J, then JG = JH = JF = HI

Given

JG - 2x+ 2

JH - 2y-4

JF-8 a

HI - 3z-3,

Hence 2x+2 = 2y-4 = 3z-3 = 8

This means that

2x+2 = 8

2x = 8-2

2x = 6

x = 3

Also, 2y-4 = 8

2y = 8+4

2y = 12

y = 6

And 3z-3 = 8

3z = 8+3

3z = 11

z = 11/3

z = 3.67 ≈ 4

Hence x = 3, y = 6,z = 4

6 0

1 year ago

Tomnhas read 11 pages of a 215 page book he will read 6 pages each day until he fi ishes the book which equation can be used to

Ierofanga [76]

Answer: 34 days

Step-by-step explanation:

To find out the amount of pages he needs to read each day at the rate of 6 pages/day is 204/6. (we divide 204 by 6 because that is 215-11, since he already read 11 pages) this is 34

5 0

7 months ago

A plane is descending towards a runway in Chicago. The plane is flying at an altitude of 7 miles and is 190 miles from the runwa

KengaRu [80]

height = 7 miles, base dimension is 190 miles

to find the angle on the ground up to the plane

use arctan(7/190)

acrtan on a calculator is tan^-1

arctan(7/190) = 2.1099 degrees. Round off as necessary

3 0

1 year ago

Cerium-144 is a radioactive isotope with a half-life of 285 days. How long would it take for an initial sample of 15 grams of Ce

serg [7]

To solve this we are going to use the formula: $A=A_{0}( \frac{1}{2} )^{ \frac{t}{h}$
where
$A$ is the final amount aster a time $t$
$A_{0}$ is the initial amount
$t$ is the time
$h$ is the half-life

We know for our problem that $A=5$ , $A_{0}=15$ , and $n=285$ , so lets replace those values in our formula:
$A=A_{0}( \frac{1}{2} )^{ \frac{t}{h}$
$5=15( \frac{1}{2} )^{ \frac{t}{285}$

Since $t$ is the exponent, we are going to use logarithms to find its value:
$5=15( \frac{1}{2} )^{ \frac{t}{285}$
$( \frac{1}{2} )^{ \frac{t}{285}}= \frac{5}{15}$
$( \frac{1}{2} )^{ \frac{t}{285}}= \frac{1}{3}$
$ln( \frac{1}{2} )^{ \frac{t}{285}}= ln(\frac{1}{3})$
$\frac{t}{285} ln( \frac{1}{2})=ln( \frac{1}{3} )$
$\frac{t}{285}= \frac{ln( \frac{1}{3}) }{ln( \frac{1}{2} )}$
$t= \frac{285ln( \frac{1}{3}) }{ln( \frac{1}{2}) }$
$t=451.71$

We can conclude that the correct answer is: a. <span>about 452 days.</span>

3 0

1 year ago