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RSB [31]
3 years ago
6

Write an equation for the transformation y=x vertical stretch by a factor of 8

Mathematics
1 answer:
Helen [10]3 years ago
7 0
y=8x\\
....................
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PLEASE HELP ILL GIVE BRAINLEST
Liono4ka [1.6K]

Answer:

4177

Step-by-step explanation:

add what rhode island is and then do the same for washington. then add both

6 0
3 years ago
(5a^2+6a+2)-(7a^2-7a-5)
krok68 [10]

For this case we must simplify the following expression:(5a^ 2 + 6a + 2) - (7a^ 2-7a-5) =

We eliminate the parentheses taking into account that:

- * + = -\\- * - = +

So:

5a ^ 2 + 6a + 2-7a ^ 2 + 7a + 5 =

We add similar terms taking into account that:

Equal signs are added and the same sign is placed.

Different signs are subtracted and the major sign is placed.

5a ^ 2-7a^2 + 6a + 7a +2 +5 =\\-2a ^ 2 + 13a + 7

Answer:

-2a ^ 2 + 13a + 7

8 0
3 years ago
What is the GCF 16 24 40
boyakko [2]
The factors of 16 are 1, 2, 4, 8, and 16. 
<span>The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. </span>
<span>The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. </span>
<span>So the greatest common factor of 16, 24, and 40 is 8.
                             
                                         Hope this helps:)</span>
4 0
2 years ago
The amount $180.00 is what percent greater than $135.00? 
sergiy2304 [10]

Answer:

Its Option C

Step-by-step explanation:

The difference is 180 - 135 = $45

So  it is 45 * 100 / 135  

= 33.33% greater

4 0
2 years ago
85.87 J or heat energy is added to a 34.8 g mass of substance the temperature rises from 21.76°C
Andru [333]

Answer:

The required specific heat is 196.94 joule per kg per °C  

Step-by-step explanation:

Given as :

The heat generated = Q = 85.87 J

Mass of substance  (m)= 34.8 gram = 0.0348 kg

Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C

Let the specific heat = S

Now we know that

Heat = Mass × specific heat × change in temperature

Or, Q = msΔt

Or, 85.87 = (0.0348 kg ) × S × 12.53°C

Or , 85.87 = 0.4360 × S

Or, S = \frac{85.87}{0.4360}

∴ S = 196.94 joule per kg per °C

Hence the required specific heat is 196.94 joule per kg per °C   Answer

7 0
2 years ago
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