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ziro4ka [17]
3 years ago
11

Consider the equation 2/5y = 20. Tell whether each statement is true or false

Mathematics
1 answer:
svlad2 [7]3 years ago
3 0

Answer:

A. True
B. False
C. False
D. True

Step-by-step explanation:

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you only have a 1/4 measuring cup and a recipe calls for 8 3/4 cups of flour how many 1/4 cups would you need to use?
Delicious77 [7]

Answer:

35 cups are needed.

Step-by-step explanation:

We have to divide 8 3/4 by 1/4.

Convert to improper fraction.

8 3/4 = [(8*4) + 3] / 4

= 35 / 4.

35/4 / 1/4

= 4*35 / 4

= 35 cups.

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3 years ago
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What is 1/2 divided by 1/2 plz show in explanation I’ll give u Brainlyist if u do
Agata [3.3K]
The answer would be 1
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3 years ago
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What is the vertex of the parabola: y=2(x-3)2^+4
denis23 [38]

Answer:

<em>Vertex: (3,4)</em>

Step-by-step explanation:

<u>Vertex form of the quadratic equation</u>

The vertex form of the quadratic function has the following equation:

y=a(x-h)^2+k

Where (h, k) is the vertex of the parabola that results when plotting the function, and a is a coefficient different from zero.

The parabola is given as:

y=2(x-3)^2+4

Comparing with the equation:

a=2, Vertex: (3,4)

The graph of the parabola is shown in the image below

5 0
3 years ago
Evaluate integral _C x ds, where C is
borishaifa [10]

Answer:

a.    \mathbf{36 \sqrt{5}}

b.   \mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}

Step-by-step explanation:

Evaluate integral _C x ds  where C is

a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)

i . e

\int  \limits _c \ x  \ ds

where;

x = t   , y = t/2

the derivative of x with respect to t is:

\dfrac{dx}{dt}= 1

the derivative of y with respect to t is:

\dfrac{dy}{dt}= \dfrac{1}{2}

and t varies from 0 to 12.

we all know that:

ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \  \ dt

∴

\int \limits _c  \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt

= \int \limits ^{12}_{0} \  \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2})  \ dt

= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0

= \dfrac{\sqrt{5}}{4}\times 144

= \mathbf{36 \sqrt{5}}

b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)

Given that:

x = t  ; y = 3t²

the derivative of  x with respect to t is:

\dfrac{dx}{dt}= 1

the derivative of y with respect to t is:

\dfrac{dy}{dt} = 6t

ds = \sqrt{1+36 \ t^2} \ dt

Hence; the  integral _C x ds is:

\int \limits _c \ x \  ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \  dt

Let consider u to be equal to  1 + 36t²

1 + 36t² = u

Then, the differential of t with respect to u is :

76 tdt = du

tdt = \dfrac{du}{76}

The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145

Thus;

\int \limits _c \ x \  ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \  dt

\mathtt{= \int \limits ^{145}_{0}  \sqrt{u} \  \dfrac{1}{72} \ du}

= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}

\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}

\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}

5 0
4 years ago
Question 1
Lera25 [3.4K]
8 units

i’m sorry if it’s wrong but i’m pretty sure it should be right.
4 0
4 years ago
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