13.15 ounces of 72% acid and 71.85 ounces of 25% acid are needed
<u>Step-by-step explanation:</u>
Total mass of acid required= 85 ounces
Let the mass of 72% acid be 'a'
Let the mass of 25% acid be 'b'
a + b = 85
b = 85-a
85(40/100) = a(72/100) + b(25/100)
85(2/5) = a(72/100) +(85- a) (25/100)
34 = (72a/100) + (2125/100) - (25a/100)
34 - (2125/100) = (72a + 25a) /100
(3400-2125)/100 = 97a /100
97a = 1275
a = 13.15 ounces
b = 85 - 13.14
b = 71.85 ounces
13.15 ounces of 72% acid and 71.85 ounces of 25% acid are needed
<span>Because the lowest two digit number is 10 and 10 • 10 = 100 so you are already into the three digit numbers and will only go higher.
i hope this help
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Using the together rate, it is found that it would take them 16 hours to sew 12 dresses.
<h3>What is the together rate?</h3>
The together rate is the <u>sum of each separate rate</u>.
In this problem, we have that:
Hence, the together rate is the time it would take them to produce one dress working together, and is given by:




Then one dress takes
of an hour, and the time for twelve dresses is:

It would take them 16 hours to sew 12 dresses.
To learn more about the together rate, you can take a look at brainly.com/question/25159431
Pemdas helps so first you do Parentheses Exponet multiply divide add and subtract
Answer:
45
Step-by-step explanation:
Two tangents drawn to a circle from an outside point form arcs and an angle, and this formula shows the relation between the angle and the two arcs.
m<EYL = (1/2)(m(arc)EVL - m(arc)EHL) Eq. 1
The sum of the angle measures of the two arcs is the angle measure of the entire circle, 360 deg.
m(arc)EVL + m(arc)EHL = 360
m(arc)EVL = 360 - m(arc)EHL Eq. 2
We are given this:
m<EYL = (1/3)m(arc)EHL Eq. 3
Substitute equations 2 and 3 into equation 1.
(1/3)m(arc)EHL = (1/2)[(360 - m(arc)EHL) - m(arc)EHL]
Now we have a single unknown, m(arc)EHL, so we solve for it.
2m(arc)EHL = 3[360 - m(arc)EHL - m(arc)EHL]
2m(arc)EHL = 1080 - 6m(arc)EHL
8m(arc)EHL = 1080
m(arc)EHL = 135
Substitute the arc measure just found in Equation 3.
m<EYL = (1/3)m(arc)EHL
m<EYL = (1/3)(135)
m<EYL = 45