Answer:
Question no.4
answer false
question no. 5
answer lifting weights
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answer doing yoga
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answer specific
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answer true
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true
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please mark brainlist answer
Answer:
y = x² - 4x - 21
Step-by-step explanation:
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
Here (h, k) = (2, - 25), thus
y = a(x - 2)² - 25
To find a substitute (7, 0) into the equation
0 = a(7 - 2)² - 25 = a(5)² - 25 = 25a - 25 ( add 25 from both sides )
25a = 25 ( divide both sides by 25
a = 1, thus
y = (x - 2)² - 25 ← in vertex form
Expand and simplify
y = x² - 4x + 4 - 25
y = x² - 4x - 21 ← in standard form
The property displayed here is the distributive property.
If you have a variable or unknown number inside or outside of parentheses, you can distribute it to each term and add the terms together, and it will remain true.
Example:
4(x + 5)
After distributing, it'll look like this:
4x + 20
To get the answer we can use proportion.
30----------100%
12-----------x
Cross multiply now
30x=12*100%
30x=1200% /:30 (divide both sides by 30)
x=40% - it's the percentage of cars which finished the race.
100%-40%=60% - it's the answer
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
