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4 years ago

Change function to standard form f(x)=(x+4)(x+5)

Mathematics

1 answer:

Sliva [168]4 years ago

6 0

Answer:

Im not sure but i think its X ^2 + 9

Step-by-step explanation:

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Ami and Dan are going to bake a cake. They need two and a half cups of flour. Ami has 5/4 cups of flour, and Dan has 4/3 cups of

Westkost [7]

Answer:

They have enough.

$2\frac{6}{12}$

Step-by-step explanation:

$\frac{4}{3}+ \frac{5}{4}$

4 = 2 × 2

3 = 3 × 1

LCM of 4 and 3

= 2 × 2 × 3 × 1

= 12

$4(\frac{4}{3})+ 3(\frac{5}{4})$

$\frac{16}{12}+ \frac{15}{12}$

$\frac{31}{12} =2\frac{7}{12}$

$2\frac{1}{2} =2\frac{6}{12}$

$2\frac{6}{12}$

6 0

3 years ago

Please help I will appreciate it, i suck at math :(

seropon [69]

Answer:

Millimeters

Step-by-step explanation:

MM is abreviation for millimeters

6 0

3 years ago

Melanie goes to a shelter to buy a dog. She has settled on to two dogs, An Alsatian and a bulldog. To decide them, she plans to

Leona [35]

52 and 58 is the answer

4 0

3 years ago

Use pythagorean theorem to find the length of the hypotenuse in the triangle shown below 11 and 60

r-ruslan [8.4K]

Answer: The value of the hypotenuse is 61.

Step-by-step explanation:

So we are given a triangle with two known sides; two legs with values of 11 and 60. We have to use the Pythagorean Theorem to find the value of the hypotenuse.

The Pythagorean Theorem states that, "the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides".

This means that:

a^2 + b^2 = c^2

leg^2 + leg^2 = hypotenuse^2

So just by looking at that, we know that we can just substitute the values in:

11^2 + 60^2 = x^2

121 + 3600 = 3721

sqrt of 3721 = 61

So the hypotenuse value is 61.

8 0

3 years ago

Write the quadratic function with the following transformations: vertical shrink by 3, left 4, down 9.

xz_007 [3.2K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see

so, the parent equation could be
y = x² ====> y = A(x + C)² + D

shrink by 3, A = 3
left 4, C = +4
down 9, D = 9

7 0

4 years ago