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nexus9112 [7]
3 years ago
6

A motorist traveled 250 miles on 11 gallons of gas. With the same vehicle, about how far could he go on 16 gallons of gas? Round

to the nearest tenth.
Mathematics
1 answer:
sesenic [268]3 years ago
6 0

Answer:

363.6

Step-by-step explanation:

250/11= 22.72

22.72 x 16= 363.3636

when rounded tot he nearest tenth its 363.6 because the three is less than five and cannot bump up the 6

if possible make my answer the brainliest

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What is 1 and 2/5× 7
OleMash [197]
7/5 x 7/1 = 49/5 or 9 4/5
3 0
3 years ago
Read 2 more answers
Carry out three steps of the Bisection Method for f(x)=3x−x4 as follows: (a) Show that f(x) has a zero in [1,2]. (b) Determine w
ELEN [110]

Answer:

a) There's a zero between [1,2]

b) There's a zero between [1.5,2]

c) There's a zero between  [1.5,1.75].

Step-by-step explanation:

We have f(x)=3^x-x^4

A)We need to show that f(x) has a zero in the interval [1, 2]. We have to see if the function f is continuous with f(1) and f(2).

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(2)=3^2-(2)^4=9-16=(-7)

We can see that f(1) and f(2) have opposite signs. And f(1)>f(2) and the function is continuous, this means that exists a real number c between the interval [1,2] where f(c)=0.

B)We have to repeat the same steps of A)

For the subinterval [1,1.5]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1) and f(1.5) have the same signs, this means there's no zero in the subinterval [1,1.5].

For the subinterval [1.5,2]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.5) and f(2) have opposite signs, this means there's a zero between the subinterval [1.5,2].

C)We have to repeat the same steps of A)

For the subinterval [1,1.25]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5

f(1) and f(1.25) have the same signs, this means there's no zero in the subinterval [1,1.25].

For the subinterval [1.25,1.5]:

f(x)=3^x-x^4\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1.25) and f(1.5) have the same signs, this means there's no zero in the subinterval [1.25,1.5].

For the subinterval [1.5,1.75]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)

f(1.5) and f(1.75) have opposite signs, this means there's a zero between the subinterval [1.5,1.75].

For the subinterval [1.75,2]:

f(x)=3^x-x^4\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.75) and f(2) have the same signs, this means there isn't a zero between the subinterval [1.75,2].

The graph of the function shows that the answers are correct.

6 0
2 years ago
HELP! please choose the correct answer from the photo
Rina8888 [55]

Answer:

10x+6

Step-by-step explanation:

the area of a rectangle is (length+height)*2

so you have to add two representations of the length and height and then multiply it by 2

7 0
2 years ago
Read 2 more answers
A. Solve the differential equation <img src="https://tex.z-dn.net/?f=y%27%3D2x%20%5Csqrt%7B1-y%5E2%7D%20" id="TexFormula1" title
kirill [66]
y' = \frac{dy}{dx}

seperable differential equations will have the form
\frac{dy}{dx} = F(x) G(y)

what you do from here is isolate all the y terms on one side and all the X terms on the other
\frac{dy}{G(y)} = F(x) dx
just divided G(y) to both sides and multiply dx to both sides

then integrate both sides
\int \frac{1}{G(y)} dy = \int F(x) dx&#10;&#10;

once you integrate, you will have a constant. use the initial value condition to solve for the constant, then try to isolate x or y if the question asks for it


In your problem,
G(y) = \sqrt{1-y^2}&#10;&#10;F(x) = 2x

so all you need to integrate is
\int \frac{1}{\sqrt{1-y^2}} dy = \int 2x dx
5 0
3 years ago
A rectangle has a length of 3x-5 and a width of 2x+9.
Montano1993 [528]

Step-by-step explanation:

l = 3x - 5

w = 2x + 9

perimeter = 2l + 2w = 2(3x - 5) + 2(2x + 9) =

= 6x - 10 + 4x + 18 = 10x + 8

area = l×w = (3x - 5)(2x + 9) = 6x² + 27x - 10x - 45 =

= 6x² + 17x - 45

8 0
2 years ago
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