4+2^2
=4+(2×2)
=4+(4)
=4+4
=8
or
(4+2)^2
=(4+2) × (4+2)
=16+8+8+4
=36
hope this helps!! please make my answer brainliest to help me out, thx!!
Answer:
Improper fraction: 39/34
Proper fraction: 1 5/34
Decimal: 1.147 (rounded to the nearest thousandths).
Step-by-step explanation:
Solve each expression given, if it has the same result, then the expressions is equivalent. Solve the given expression:
4 7/8 ÷ 4 1/4
First, change all fractions into improper fractions:
4 7/8 = (4 * 8)/8 + 7/8 = 32/8 + 7/8 = 39/8
4 1/4 = (4 * 4)/4 + 1/4 = 16/4 + 1/4 = 17/4
Next, find common denominators. Note that what you do to the denominator, you must do the numerator:
(17/4) * (2/2) = (34/8)
39/8 ÷ 17/4
Solve. First, change the division sign into multiplication, and then flip the second fraction:
39/8 ÷ 17/4 = 39/8 x 4/17
Multiply straight across:
39/8 x 4/17 = (39 x 4)/(8 x 17) = 156/136 = 39/34 (simplified), or 1 5/34.
~
Answer:
see below
Step-by-step explanation:
<h3>Proposition:</h3>
Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE
<h3>Proof:</h3>
1)The lines AD and BC are parallel and AC their transversal therefore,
![\displaystyle \angle DAC = \angle ACB \\ \ \qquad [\text{ alternate angles theorem}]](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Cangle%20DAC%20%3D%20%20%5Cangle%20ACB%20%5C%5C%20%20%5C%20%5Cqquad%20%5B%5Ctext%7B%20alternate%20angles%20theorem%7D%5D)
2)The lines AB and DC are parallel and BD their transversal therefore,
![\displaystyle \angle BD C= \angle ABD \\ \ \qquad [\text{ alternate angles theorem}]](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Cangle%20BD%20C%3D%20%20%5Cangle%20ABD%20%5C%5C%20%20%5C%20%5Cqquad%20%5B%5Ctext%7B%20alternate%20angles%20theorem%7D%5D)
3)now in triangle ∆AEB and ∆CED
therefore,
hence,
Proven
Answer:
slope is 2
y int is (0,7)
Step-by-step explanation:
