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Yuliya22 [10]
3 years ago
15

X^2+17x=-42 how do i find the solutions to this problem?

Mathematics
1 answer:
ELEN [110]3 years ago
3 0

photo math duhhhhhhhhhhhhhhhhhhhh

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Please answer im wasting pointss!!!!
kondor19780726 [428]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Write an equation of a parabola that opens to the left, has a vertex at the origin, and a focus at (–4, 0).
8_murik_8 [283]

Answer:

y^{2}=-16x

Step-by-step explanation:

we know that

The standard equation of a horizontal parabola is equal to  

(y-k)^{2}=4p(x-h)

where

(h,k) is the vertex

(h+p,k) is the focus

In this problem we have

(h,k)=(0,0) ----> vertex at origin

(h+p,k)=(-4,0)

so

h+p=-4

p=-4

substitute the values

(y-0)^{2}=4(-4)(x-0)

y^{2}=-16x

8 0
3 years ago
How do I solve this:<br> (-3z2+z+13)-(6z2+7)
vesna_86 [32]
I know that you should use PEMDAS, what grade are you in this sems advances for you
7 0
3 years ago
John needs to make a total of 35 deliveries this week. So far he has completed 28 of them. What percentage of his total deliveri
lukranit [14]

Answer:

80%

Step-by-step explanation:

28/35=4/5

4/5=0.80

0.80=80%

6 0
2 years ago
Please help, performance task: trigonometric identities
AnnZ [28]

The solutions to 1 - cos(x) = 2 - 2sin²(x) from (-π, π) are (-π/3, 0.5) and (π/3, 0.5)

<h3>How to solve the trigonometric equations?</h3>

<u>Equation 1: 1 - cos(x) = 2 - 2sin²(x) from (-π, π)</u>

The equation can be split as follows:

y = 1 - cos(x)

y = 2 - 2sin²(x)

Next, we plot the graph of the above equations (see graph 1)

Under the domain interval (-π, π), the curves of the equations intersect at:

(-π/3, 0.5) and (π/3, 0.5)

Hence, the solutions to 1 - cos(x) = 2 - 2sin²(x) from (-π, π) are (-π/3, 0.5) and (π/3, 0.5)

<u>Equation 2: 4cos⁴(x) - 5cos²(x) + 1 = 0 from [0, 2π)</u>

The equation can be split as follows:

y = 4cos⁴(x) - 5cos²(x) + 1

y = o

Next, we plot the graph of the above equations (see graph 2)

Under the domain interval [0, 2π), the curves of the equations intersect at:

(π/3, 0), (2π/3, 0), (π, 0), (4π/3, 0) and (5π/3, 0)

Hence, the solutions to 4cos⁴(x) - 5cos²(x) + 1 = 0 from [0, 2π) are (π/3, 0), (2π/3, 0), (π, 0), (4π/3, 0) and (5π/3, 0)

Read more about trigonometry equations at:

brainly.com/question/8120556

#SPJ1

4 0
1 year ago
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