2) angle T Congruent to angle D; 3)angle s and angle u
Answer:
Let y represents the ounces of first spice.
From the given statements, we draw a table as shown below:
ounces of spice Percentage ounces of salt
First spice y 3% 0.03x
Second spice 175 6% 175(0.06)
Final mixture y + 175 5% (x+15)(0.05)
Now, to solve for y
we have;
Final spice = First spice + second spice


Subtract 10.5 on both sides we have;

Simplify:

Subtract 0.05y on both sides

Simplify:

Divide both sides by -0.02, we get;

Therefore, 87.5 ounces of spices that is 3% salt should be added to 175 ounces of a spice that is 6% salt in order to make a spice that is 5% salt
Answer:
Hi there!
The answer to this question is c=23, b and d=157
this is because c is the vertical angle of a meaning the values are equal
and b is the supplementary angle to a: all you do is subtract 23 from 180 to get 157.
b is vertical to d, so d is also 157
Choice A is the answer which is the point (1,-1)
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How I got this answer:
Plug each point into the inequality. If you get a true statement after simplifying, then that point is in the solution set and therefore a solution. Otherwise, it's not a solution.
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checking choice A
plug in (x,y) = (1,-1)



This is true because -3 is equal to itself. So this is the answer.
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checking choice B
plug in (x,y) = (2,4)



This is false because 0 is not to the left of -3, nor is 0 equal to -3. We can cross this off the list.
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checking choice C
plug in (x,y) = (-2,3)



This is false because 7 is not to the left of -3, nor is 7 equal to -3. We can cross this off the list.
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checking choice D
plug in (x,y) = (3,4)



This is false because -2 is not to the left of -3, nor is -2 equal to -3. We can cross this off the list.
Answer:
60 chickens
Step-by-step explanation:
We assume the 140 refers to the number animals (heads, noses, bodies, whatever). If all of those were chickens, there would be 280 legs. There are 160 legs more than that.
Since each pig that replaces a chicken adds 2 legs, there must be 160/2 = 80 pigs. That leaves 140 -80 = 60 chickens.