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ICE Princess25 [194]
3 years ago
15

Someone plz help me I’ll give brainliest

Mathematics
1 answer:
True [87]3 years ago
3 0

Answer:

f(x)= 2x- 2h

Step-by-step explanation:

so your answer is 2 :)

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-5.5x+0.77=1.48 whats x
Nesterboy [21]

Answer:

X= -0.1290 (bar line on top of the 90)

Hope this helps!

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
The two-way frequency table below shows data on student behavior and the use of positive phone calls home as an incentive for go
dusya [7]

Answer:

From the frequency table, let's calculate the row total.

Row total for phone call = 19 + 9 = 28

Row total for no phone call = 8 +6 = 14

To calculate their respective row relative frequencies, let's use:

Row relative freq = \frac{freq.}{Row total}

Now, the two-way frequency table will be computed as:

For phone call:

Desirable behavior = \frac{19}{28} = 0.67857 ≈0.69

Undesirable behaviour = \frac{9}{28} = 0.3214 ≈0.32

No phone call:

Desirable behaviour = \frac{8}{14} = 0.5714 ≈ 0.57

Undesirable behaviour = \frac{6}{14} = 0.4286 ≈ 0.43

The complete two-way table is attached.

8 0
3 years ago
find the dimensions of a rectangle whose width is 5 miles less than its length and whose area is 66 square miles
Lyrx [107]
A = l * w
66 = l * (l - 5)
66 = l^{2}-5l
0 = l^{2}-5l - 66

l = 11 or 6 check which works.

A = 11 * 6, which fits.

length = 11 miles
width = 5 miles
4 0
3 years ago
4x^2-y^2+80x+16y+272=0
Andrew [12]

Answer:

\frac{4x + 40}{-y +8}

Step-by-step explanation:

8 0
3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
Elden [556K]
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.

Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.

(i)

V=\pi\cdot\int\limits_a^bf^2(x)\, dx\qquad\implies \qquad a=0\qquad b=9\qquad f(x)=x^3


V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}

Answer B. or D.

(ii)

V=2\pi\cdot\int\limits_a^bx\cdot f(x)\, dx


V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx=
2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}

So we know that the correct answer is D.

(iii)
Line x = h

V=2\pi\cdot\int\limits_a^b(h-x)\cdot f(x)\, dx\qquad\implies\qquad h=18


V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\

=2\pi\cdot\left(18\cdot\dfrac{9^4}{4}-\dfrac{9^5}{5}\right)=2\pi\cdot\dfrac{177147}{10}=\boxed{\dfrac{177147\pi}{5}}

Answer D. just as before.

6 0
3 years ago
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