Answer:
D) 3x^2 - 12
Step-by-step explanation:
Using PEMDAS;
There is no need to evaluate the part of the equation (x^2 - 8) because is no need to, as it is already in its simplest form.
We must evaluate the part of the equation continuing with, "- (-2x^2+4)," as it is not in its simplest form.
Evaluating "- (-2x^2+4)":
Step 1: Distributing the negative
Once distributing the negative symbol amongst the values within the parenthesis according to PEMDAS, we get "2x^2 - 4" as the product.
Step 2: Consider the rest of the equation to evaluate
Since the part of the equation is still in play here as it is a part of the original equation to be solved, we must evaluate it as a whole to get the final answer.
Thus,
x^2 -8 + 2x^2 - 4 = ___
*we can remove the parenthesis as it has no purpose, since it makes no difference.
Evaluating for the answer, we get,
x^2+2x^2 + (-8 - 4) = 3x^2 - 12
Hence, the answer is D) 3x^2 - 12.
Answer:
Step-by-step explanation:
Find the LCD because like last time, you cannot add or subtract fractions with different denominators.
LCD = 10
Since the second fractions stayed at the same denominator, you will only have to multiply the first fraction.
Set the equations equal to each other
Subtract x from both sides
Divide both sides by -4
x=-1/2
1. Angles ADC and CDB are supplementary, thus
m∠ADC+m∠CDB=180°.
Since m∠ADC=115°, you have that m∠CDB=180°-115°=65°.
2. Triangle BCD is isosceles triangle, because it has two congruent sides CB and CD. The base of this triangle is segment BD. Angles that are adjacent to the base of isosceles triangle are congruent, then
m∠CDB=m∠CBD=65°.
The sum of the measures of interior angles of triangle is 180°, therefore,
m∠CDB+m∠CBD+m∠BCD=180° and
m∠BCD=180°-65°-65°=50°.
3. Triangle ABC is isosceles, with base BC. Then
m∠ABC=m∠ACB.
From the previous you have that m∠ABC=65° (angle ABC is exactly angle CBD). So
m∠ACB=65°.
4. Angles BCD and DCA together form angle ACB. This gives you
m∠ACB=m∠ACD+m∠BCD,
m∠ACD=65°-50°=15°.
Answer: 15°.
I did the math,This is false.
Hope this helps.
