How do I put -(x-4)^2 -1 into standard form

1 answer:

4 0

Standard form is ax²+bx+c

expand then simlify
-(x-4)²=-(x-4)(x-4)=-(x²-8x+16)=-x²+8x-16
now we gots
-x²+8x-16-1
-x²+8x-17 is standard form

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What is negative 21/9 times negative 18/7

diamong [38]

Answer:

negative 21/9 times negative 18/7 is 6

Step-by-step explanation:

The phrase negative 21/9 times negative 18/7 can be written mathematically as

$-\frac{21}{9} \times -\frac{18}{7}$

Lets first convert the fractions into decimal values

$-\frac{21}{9}$ = -2.33

$-\frac{18}{7}$ = -2.57

Now on simplyfying

=> $-2.33 \times -2.57$

=> 5.988

=>6 (rounding off)

We will get a positive value because the multiplicative law states that the product of two negative number is positive

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3 years ago

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egoroff_w [7]

Answer:

Solution given:

r=3km

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is a required answer

6 0

3 years ago

Read 2 more answers

A Find tan a. r √5,-√7) ✓[? Enter

Virty [35]

Answer:

$tan(\alpha)=-\frac{\sqrt{35}}{5}$

Step-by-step explanation:

Tan can be defined as: $\frac{sin(\theta)}{cos(\theta)}$ as it simplifies to opposite/adjacent. If you know a bit about the unit circle, you'll know that the x-coordinate is going to be cos(theta) and the y-coordinate is going to be sin(theta). Since the sin(theta) is defined as opposite/hypotenuse, and the hypotenuse = 1, so sin(theta) is defined as the opposite side, which is the y-axis. Same thing goes for cos(theta), except the adjacent side is the x-axis.

Using this we can define tan

$sin(\alpha)=-\sqrt{7}\\cos(\alpha)=\sqrt{5}\\\\tan(\alpha)=-\frac{\sqrt{7}}{\sqrt{5}} * \frac{\sqrt{5}}{\sqrt{5}}\\tan(\alpha)=-\frac{\sqrt{7*5}}{5}\\tan(\alpha)=-\frac{\sqrt{35}}{5}\\$