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2 years ago

Negative ten is no less than two times a number plus fourteen.

Mathematics

1 answer:

masha68 [24]2 years ago

3 0

Answer:

i this like true or false,if so its true

Step-by-step explanation:

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What is the interval of increase and decrease of f(x)= (x-4)^2 -1

nevsk [136]

Answer:

(−∞,4)

Step-by-step explanation:

Explanation:

To determine the interval that f(x) is increasing.

increasing when f'(x)>0f'(x)=−2(x−4)← using chain rule

solve −2(x−4)>0

⇒−2x+8>0

⇒x<4

interval is (−∞,4)

graph{-(x-4)^2-3 [-8.89, 8.89, -4.445, 4.44]}

3 0

2 years ago

Please help!!!!!!!!!!!!!!!

natita [175]

The slope is -8/3, because the slope is always before the x, and is the coefficient. The coefficient is 4.

7 0

2 years ago

Determine the signs of the quotients.

Novosadov [1.4K]

Answer:

The quotient of -125 and 5 is : -25, negative

The quotient of -264 and -12 is : 22 positive

The quotient of O and -350 is : 0 neither negative nor positive

The quotient of 276 and 8 is : 36 positive

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Find the sum of 1/6and 7/12 in lowest terms.

Nady [450]

Answer:

11/12

Step-by-step explanation:

1/6 multiply both my 2 to get the same denominator of 12. it would be 4/12 then add 7/12 then you'll get 11/12.

5 0

2 years ago

Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).

Burka [1]

Step-by-step explanation:

As we know that

The circle center is equidistant from all three points, the distance being the circle radius.

Any point equidistant from two points must lie on the perpendicular bisector of the line segment which join those two points.

Which is, on the line through the midpoint of the line segment, perpendicular to the line segment.

The perpendicular bisector of the line segment joining the points (1, 1) and (1, 3) will be:

$\:y=\:\frac{1+3}{2}=\frac{4}{2}=2$

The perpendicular bisector of the line segment joining the points (1, 3), and (9, 2) will be:

$x=\:\frac{1+9}{2}=\frac{10}{2}=5$

These intersect at the center of the circle (5, 2).

The distance between (1, 1) and (5, 2) will be:

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\mathrm{The\:distance\:between\:}\left(1,\:1\right)\mathrm{\:and\:}\left(5,\:2\right)\mathrm{\:is\:}$

$=\sqrt{\left(5-1\right)^2+\left(2-1\right)^2}$

$=\sqrt{4^2+1}$

$=\sqrt{16+1}$

$=\sqrt{17}$

So the equation of the circle can be written as:

$\left(x-5\right)^2+\left(y-2\right)^2=17$

$x^2-10x+y^2+29-4y=17$

$x^2-10x+y^2-4y+12\:=0$

$x^2+y^2-10x-4y+12=0$

8 0

2 years ago