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valentina_108 [34]
4 years ago
7

According to the Rational Roots Theorem, which statement about f(x) = 25x7 – x6 – 5x4 + x – 49 is true?

Mathematics
2 answers:
kvv77 [185]4 years ago
8 0
F(x)=20x-52 so it would be 2.6
ad-work [718]4 years ago
3 0

Answer:

C. Any rational root of f(x) is a factor of –49 divided by a factor of 25.

Step-by-step explanation:

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Write an algebraic expression for this situation . Steve made a down payment of $300 on a new tv. He paid $37 per month for x mo
Temka [501]
300+37x
300+37(9)
300+333
633
5 0
4 years ago
Please help me solve this equation. 15 points ASAP!!
coldgirl [10]

Answer:

The answer is

x=-1 or x=4

3 0
4 years ago
Lisa and Bart spin this spinner 60 times the result are below/ black=17 blue=15 orange=21 purple=7 a. What is the experimental p
asambeis [7]

\text{Answer: a) }P(\text{ getting an orange )}=\frac{7}{20}

b) Blue color had an experimental probability that matched its theoretical probability.

Explanation:

Since we have given that

Number of times this spinner is spinned = 60

Number of times black occur = 17

Number of times blue occur = 15

Number of times orange occur = 21

Number of times purple occur = 7

a) So, Experimental probability of a spin of orange is given by

P(\text{ getting an orange }=\frac{\text{ Number of times orange occur}}{\text{total number of times the spinner spins}}\\\\P(\text{ getting an orange})}=\frac{21}{60}=\frac{7}{20}

b) which color had an experimental probability that matched its theoretical probability.

According to theoretical probability ,

Every event must have equal probability, i.e. \frac{1}{4}

And,

P(\text{ getting blue color)}=\frac{15}{60}=\frac{1}{4}

So, Blue color had an experimental probability that matched its theoretical probability.





7 0
3 years ago
Read 2 more answers
Someone help please?! Need right answers only don’t come for the points will mark brainliest
Kaylis [27]
Hi there!

These can probably be done on your own. You just gotta know what to do! :)

Let's take #1 for example. You (or maybe a classmate/teacher showed you?) plotted the points. Mark each point with the given letter, so you don't get lost. Then, you reflected it over the y-axis.

Think of it as a mirror. Say you held a picture of a rhombus up to it. You would see the rhombus, yourself, and whatever was in the background reflected back at you. You step closer, the image steps closer. You turn the rhombus, and the image also turns. This principle can be used here!

So, keep doing what you're doing. Here's a step-by-step:

1.) Plot each point, and mark its name. For example, 'B' is (-6,7), and you write 'B' next to the point.

2.) Double check the point are exactly where they need to be

3.) Connect each point with a straight line. You can use a ruler, student ID, whatever as a straightedge, but it looks neater

4.) Draw a line for the axis. For example, if y=0, draw a straight line again there. (hint: that's the y-axis!)

5.) Double check that everything is right so far again. This is easy to mess up!

6.) Reflect each point over the axis. Another example, (-3, 2) becomes (3, 2). Mark this with an apostrophe (') to signal the point as prime, or the reflected point. For example, B becomes B' (B prime)

7.) Check one final time

If you found this especially helpful, I'd appreciate if you'd vote me Brainliest for your answer. I want to be able to assist more users one-on-one! :)
6 0
3 years ago
Solve the initial value problem y′+y=f(t),y(0)=0 where f(t)={1,−1, if t<4 if t≥4 Use h(t−a) for the Heaviside function shifte
Anton [14]

Looks like the function on the right hand side is

f(t)=\begin{cases}1&\text{for }t

We can write it in terms of the Heaviside function,

h(t-a)=\begin{cases}1&\text{for }t\ge a\\0&\text{for }t>a\end{cases}

as

f(t)=h(t)-2h(t-4)

Now for the ODE: take the Laplace transform of both sides:

y'(t)+y(t)=f(t)

\implies s Y(s)-y(0)+Y(s)=\dfrac{1-2e^{-4s}}s

Solve for <em>Y</em>(<em>s</em>), then take the inverse transform to solve for <em>y</em>(<em>t</em>):

(s+1)Y(s)=\dfrac{1-e^{-4s}}s

Y(s)=\dfrac{1-e^{-4s}}{s(s+1)}

Y(s)=(1-e^{-4s})\left(\dfrac1s-\dfrac1{s+1}\right)

Y(s)=\dfrac1s-\dfrac{e^{-4s}}s-\dfrac1{s+1}+\dfrac{e^{-4s}}{s+1}

\implies y(t)=1-h(t-4)-e^{-t}+e^{-(t-4)}h(t-4)

\boxed{y(t)=1-e^{-t}-h(t-4)(1-e^{-(t-4)})}

4 0
3 years ago
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