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Anastaziya [24]
3 years ago
12

Quick help please! I can't figure out the answer, I got this same question wrong 2 times in a row.. Any help is appreciated!

Mathematics
1 answer:
dybincka [34]3 years ago
6 0
It has one X-intercept. Graph it on Desmos and you can see the line, and then it becomes obvious it is one x-intercept.

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Determine the standard form of the equation for the circle with center (h,k) = (-5 ½) and radius r =1
wariber [46]
Asuuing yo meant (h,k)=(-5,1/2)

equation is (x-h)^2+(y-k)^2=r^2
(h,k) is center and r=radious
given that (h,k)=(-5,1/2) and r=1
h=-5
k=1/2

(x-(-5))^2+(y- \frac{1}{2} )^2=1^2
(x+5)^2+(y- \frac{1}{2} )^2=1
if you wanted us to expand
x²+y^2+10x-y+25.25=1
8 0
3 years ago
Andrea constructed a triangle. Angle 1 and 3 are the same size and angle 2 has a measurement of 70 degrees. What is the measurem
Verdich [7]

Answer:

Step-by-step explanation:

By triangle sum theorem,

Sum of all angles of a triangle is 180°.

m∠1 + m∠2 + m∠3 = 180°

(m∠1 + m∠3) + m∠2 = 180°

2(m∠1) + 70° = 180° {Given → m∠1 = m∠3]

2(m∠1) = 110°

m∠1 = 55°

Therefore, m∠1 = m∠3 = 55°

7 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!!
Dmitriy789 [7]

Answer:

7.11\leq-7.1

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is 41.073 rounded to 0
lora16 [44]
41.1 is the answer to this question.
4 0
4 years ago
If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, find the rate (in cm/min) at which the diameter d
grin007 [14]

Answer:

The diameter decreases at a rate of 0.053 cm/min.

Step-by-step explanation:

Surface area of an snowball

The surface area of an snowball has the following equation:

S_{a} = \pi d^2

In which d is the diameter.

Implicit differentiation:

To solve this question, we differentiate the equation for the surface area implictly, in function of t. So

\frac{dS_{a}}{dt} = 2d\pi\frac{dd}{dt}

Surface area decreases at a rate of 3 cm2/min

This means that \frac{dS_{a}}{dt} = -3

Tind the rate (in cm/min) at which the diameter decreases when the diameter is 9 cm.

This is \frac{dd}{dt} when d = 9. So

\frac{dS_{a}}{dt} = 2d\pi\frac{dd}{dt}

-3 = 2*9\pi\frac{dd}{dt}

\frac{dd}{dt} = -\frac{3}{18\pi}

\frac{dd}{dt} = -0.053

The diameter decreases at a rate of 0.053 cm/min.

7 0
3 years ago
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