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Leya [2.2K]
2 years ago
8

Factor completely 4n^2-28n+49

Mathematics
2 answers:
svetlana [45]2 years ago
6 0

Answer:

Then it would be 4n squared + 21

Step-by-step explanation:

topjm [15]2 years ago
6 0

Answer:

(2n - 7)^2

Step-by-step explanation:

both 4n^2 and 49 are perfect squares so take their square roots and get 2n and 7

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The total expenditures on benefits​ (in billions of​ dollars) can be approximated by the function h(x)=23.5(1.08)x​, where x​ =
Katyanochek1 [597]
Sorry don’t know but good luck
5 0
3 years ago
deon earned a total of $61 from selling jewelry at a fair. He earned $5 for selling one necklace. He also earned $8 for each bra
Elis [28]

Answer: 7

Step-by-step explanation:

8x +5 = 61

subtract 5 from both sides

8x = 56 divide both sides by 8 to get x by itself

x= 7

7 0
3 years ago
Is the following relation a function?<br> ху<br> 1 -2<br> 1-3<br> 2 1<br> 3 - 2<br> Yes<br> No
professor190 [17]

Answer:

no

Step-by-step explanation:

In order to be a function you should have one output for every input (or one y for every x). In this case there are multiple outputs for the number one which means the relationship isn't a function

4 0
3 years ago
Hi can you help me with this problem i dont understand it​
Romashka-Z-Leto [24]

Answer:

Answer is d - x = \frac{y + z}{4}

Step-by-step explanation:

Step 1: Add 'z' to both sides

4x -z (+z) = y + z

4x = y + z

Step 2: To solve for x, divide both sides by 4

\frac{4x}{4} = \frac{y + z}{4}

x = \frac{y + z}{4}

5 0
3 years ago
Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
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