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Nimfa-mama [501]
3 years ago
14

Help plzzzz thanks guys for hleping

Mathematics
1 answer:
beks73 [17]3 years ago
4 0
What does h say? The photo is cut off
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The temperature is 50 degrees Fahrenheit. The temperature will decrease by 4 degrees F. each hour. Let h be the number of hours.
goblinko [34]

The correct answer choice is B

8 0
3 years ago
Read 2 more answers
Question:
prohojiy [21]
Let f(x) = p(x)/q(x), where p and q are polynomials and reduced to lowest terms. (If p and q have a common factor, then they contribute removable discontinuities ('holes').) 
Write this in cases: 
(i) If deg p(x) ≤ deg q(x), then f(x) is a proper rational function, and lim(x→ ±∞) f(x) = constant. 
If deg p(x) < deg q(x), then these limits equal 0, thus yielding the horizontal asymptote y = 0. 
If deg p(x) = deg q(x), then these limits equal a/b, where a and b are the leading coefficients of p(x) and q(x), respectively. Hence, we have the horizontal asymptote y = a/b. 
Note that there are no obliques asymptotes in this case. ------------- (ii) If deg p(x) > deg q(x), then f(x) is an improper rational function. 
By long division, we can write f(x) = g(x) + r(x)/q(x), where g(x) and r(x) are polynomials and deg r(x) < deg q(x). 
As in (i), note that lim(x→ ±∞) [f(x) - g(x)] = lim(x→ ±∞) r(x)/q(x) = 0. Hence, y = g(x) is an asymptote. (In particular, if deg g(x) = 1, then this is an oblique asymptote.) 
This time, note that there are no horizontal asymptotes. ------------------ In summary, the degrees of p(x) and q(x) control which kind of asymptote we have. 
I hope this helps!
4 0
3 years ago
Or a two-sided confidence interval with 13 observations, σx unknown, and α = 0.10 the (critical) value is:
gavmur [86]
This is pre calc so it is 0.233
7 0
3 years ago
If Mary pay $3695.20for principal and interest every month for 30 years on her $110,000 loan, how much interest will she pay ove
PilotLPTM [1.2K]

Answer:

  $1,220,200

Step-by-step explanation:

The total of Mary's payments is ...

  $3695.20/mo × 30 yr × 12 mo/yr = $1,330,200

The difference between this repayment amount and the value of her loan is the interest she pays:

  $1,330,200 -110,000 = $1,220,200 . . . total interest paid

_____

Mary's effective interest rate is about 40.31% per year--exorbitant by any standard.

6 0
3 years ago
Find the area of a triangle bounded by the y-axis, the line f(x)=9−4/7x, and the line perpendicular to f(x) that passes through
Setler79 [48]

<u>ANSWER:  </u>

The area of the triangle bounded by the y-axis is  \frac{7938}{4225} \sqrt{65} \text { unit }^{2}

<u>SOLUTION:</u>

Given, f(x)=9-\frac{-4}{7} x

Consider f(x) = y. Hence we get

f(x)=9-\frac{-4}{7} x --- eqn 1

y=9-\frac{4}{7} x

On rewriting the terms we get

4x + 7y – 63 = 0

As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.

Area of right angle triangle = \frac{1}{ab} where a, b are lengths of sides other than hypotenuse.

So, we need find length of f(x) and its perpendicular line.

First let us find perpendicular line equation.

Slope of f(x) = \frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}

So, slope of perpendicular line = \frac{-1}{\text {slope of } f(x)}=\frac{7}{4}

Perpendicular line is passing through origin(0,0).So by using point slope formula,

y-y_{1}=m\left(x-x_{1}\right)

Where m is the slope and \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)

y-0=\frac{7}{4}(x-0)

y=\frac{7}{4} x --- eqn 2

4y = 7x

7x – 4y = 0  

now, let us find the vertices of triangle, one of them is origin, second one is point of intersection of y-axis and f(x)

for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

7y = 63

y = 9

Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}

Put x value in (2)

\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}

So, the point of intersection is \left(\frac{252}{65}, \frac{441}{65}\right)

Length of f(x) is distance between \left(\frac{252}{65}, \frac{441}{65}\right) and (0,9)

\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}

Now, length of perpendicular of f(x) is distance between \left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)

\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}

Now, area of right angle triangle = \frac{1}{2} \times \frac{252}{65} \times \frac{63}{65} \sqrt{65}

=\frac{7938}{4225} \sqrt{65} \text { unit }^{2}

Hence, the area of the triangle is \frac{7938}{4225} \sqrt{65} \text { unit }^{2}

8 0
3 years ago
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