60 miles because (20mph)((3hrs) = 60
Has 16 cups of water that is half full.
16/2 = 8
Watering can then has 8 cups of water in it.
This number needs to be divided by 15 plants.
The answer is 8/15 of a cup of water per plant which is also 0.53 cups per plant.
Given:
The vertices of a polygon QRST are T(-2, 3), Q(1, 5), R(3, -1) and S(0, 0).
To find:
The vertices for 
.
Solution:
The rule represents refection of polygon QRST across the x-axis.
If a figure is reflected across the x-axis, then
Using this rule, we get
Therefore, the vertices for are T'(-2, -3), Q'(1, -5), R'(3, 1) and S'(0, 0).
The given expression 2^8 * 8^2 * 4^-4 can be written in the exponential form 2^n as 2^6.
<h3>What are exponential forms?</h3>
The exponential form is a more convenient way to write repetitive multiplication of the same integer by using the base and its exponents.
<u>For example:</u>
If we have a*a*a*a, it can be written in exponential form as:
=a^4
where
- a is the base, and
- 4 is the power.
The power in this format reflects the number of times we multiply the base by itself. The exponent is also known as the index or power.
From the information given:
We can write 2^8 * 8^2 * 4^-4 in form of 2^n as follows:
Therefore, we can conclude that by using the exponential form, the given expression 2^8 * 8^2 * 4^-4 in the form 2^n is 2^6.
Learn more about exponential forms here:
brainly.com/question/8844911
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This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
