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Digiron [165]
3 years ago
10

What is the domain and range of f(x)=4^x-8

Mathematics
1 answer:
gladu [14]3 years ago
8 0

If the function is: f(x) = 4^x     - 8

D = {XeR}
R = {YeR|Y≥0}


if the function is: <span>f(x)=4^(x-8)

</span>D = {XeR}
R = {YeR|Y≥ -8}
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In un trapezio rettangolo la base minore, il lato obliquo e l'altezza misurano rispettivamente 60 cm. 95 cm è 76 cm. Calcola il
KengaRu [80]

Answer:

2p=348 cm\: S=6726 cm^{2}

Step-by-step explanation:

Ciao, come stai?

1) Per prima cosa, dobbiamo trovare la misura della base più grande. Scomponendo la figura possiamo visualizzare un triangolo e un quadrato. Ci sono somiglianze con gli angoli. Quindi è un triangolo rettangolo. Applichiamo il teorema di Pitagora:

a^2=b^2+c^2\\95^2=b^2+76^2\\57=b

2) Perimetro:

2p=60+57+76+60+95\\2p=348 cm

3) L' area

\frac{(B+b)h}{2} =\frac{(117+60)76}{2} =6726 \:cm^{2}

8 0
3 years ago
How many total hours did the girls spend cleaning their home
solmaris [256]

Answer:

it should be 12

Step-by-step explanation:

because 3x4=12

7 0
2 years ago
Read 2 more answers
To lose one pound of fat, a 200-pound person must burn 3,500 calories. If that person burns 180 calories by walking for 30 minut
SVEN [57.7K]

Answer:

C - 19 hours 27 minutes

Step-by-step explanation:

180 × 2 = 360

360 calories in an hour

3500 × 2 = 7000 (amount of calories in 2 pounds)

7000 ÷ 360 = 19.4444...


I hope this helps


4 0
3 years ago
Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6
Anton [14]

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

V= \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder,x^{2}+y^{2} =1 and the ellipsoid, 4x^{2}+ 4y^{2} + z^{2} =64

In polar coordinates, we know that

x^{2}+y^{2} =r^{2}

So, the ellipsoid gives

4{(x^{2}+ y^{2)} + z^{2} =64

4(r^{2}) + z^{2} = 64

z^{2} = 64- 4(r^{2})

z=± \sqrt{64-4r^{2} }

So, the volume of the solid is given by:

V= \int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta

= 2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta

To solve the integral take, 64-4r^{2} = t

dt= -8rdr

rdr = \frac{-1}{8} dt

So, the integral  \int\ r\sqrt{64-4r^{2} } rdr become

=\int\ \sqrt{t } \frac{-1}{8} dt

= \frac{-1}{12} t^{3/2}

=\frac{-1}{12} (64-4r^{2}) ^{3/2}

so on applying the limit, the volume becomes

V= 2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta

=\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta

V = \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Since, further the integral isn't having any term of \theta.

we will end here.

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

#SPJ4

3 0
1 year ago
The equation gives the speed at impact, V metres per second, of an object dropped from a height of h metres. SHOW WORK From what
devlian [24]

Answer:

h = 17.65 m

Step-by-step explanation:

The given equation gives the speed at impact :

v=\sqrt{2gh}

h is height form where the object is dropped

Put v = 18.6 m/s in the above equation.

h=\dfrac{v^2}{2g}\\\\h=\dfrac{(18.6)^2}{2\times 9.8}\\\\h=17.65\ m

So, the object must be dropped from a height of 17.65 m.

4 0
3 years ago
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