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Svetllana [295]
3 years ago
15

Nine tenths of a sum a money 2.70find the sum of money

Mathematics
1 answer:
irakobra [83]3 years ago
7 0

Answer:

$3.00

Step-by-step explanation:

Setting up an equation with a variable and using key words from the problem will allow you to solve for the sum of money.

'x' = sum money

'nine tenths of a sum' = \frac{9}{10}x

'is 2.70':  = 2.70

\frac{9}{10}x=2.70

Multiply both sides by the reciprocal of 9/10 (10/9):  

(10/9)(9/10)x = 2.70(10/9)

x = 27/9 = $3.00

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What is the area of the shaded region?​
IRINA_888 [86]
20 because of the whole is 40 and the shaded part is by 20 you just take half of 40 and you get 20
3 0
3 years ago
A)A cuboid with a square x cm and height 2xcm². Given total surface area of the cuboid is 129.6cm² and x increased at 0.01cms-¹.
Nutka1998 [239]

Answer: (given assumed typo corrections)


(V ∘ X)'(t) = 0.06(0.01t+3.6)^2 cm^3/sec.


The rate of change of the volume of the cuboid in change of volume per change in seconds, after t seconds. Not a constant, for good reason.



Part B) y'(x+Δx/2)×Δx gives exactly the same as y(x+Δx)-y(x), 0.3808, since y is quadratic in x so y' is linear in x.


Step-by-step explanation:

This problem has typos. Assuming:

Cuboid has square [base with side] X cm and height 2X cm [not cm^2]. Total surface area of cuboid is 129.6 cm^2, and X [is] increas[ing] at rate 0.01 cm/sec.


129.6 cm^2 = 2(base cm^2) + 4(side cm^2)

= 2(X cm)^2 + 4(X cm)(2X cm)

= (2X^2 + 8X^2)cm^2

= 10X^2 cm^2

X^2 cm^2 = 129.6/10 = 12.96 cm^2

X cm = √12.96 cm = 3.6 cm


so X(t) = (0.01cm/sec)(t sec) + 3.6 cm, or, omitting units,

X(t) = 0.01t + 3.6

= the length parameter after t seconds, in cm.


V(X) = 2X^3 cm^3

= the volume when the length parameter is X.


dV(X(t))/dt = (dV(X)/dX)(X(t)) × dX(t)/dt

that is, (V ∘ X)'(t) = V'(X(t)) × X'(t) chain rule


V'(X) = 6X^2 cm^3/cm

= the rate of change of volume per change in length parameter when the length parameter is X, units cm^3/cm. Not a constant (why?).


X'(t) = 0.01 cm/sec

= the rate of change of length parameter per change in time parameter, after t seconds, units cm/sec.

V(X(t)) = (V ∘ X)(t) = 2(0.01t+3.6)^3 cm^3

= the volume after t seconds, in cm^3

V'(X(t)) = 6(0.01t+3.6)^2 cm^2

= the rate of change of volume per change in length parameter, after t seconds, in units cm^3/cm.

(V ∘ X)'(t) = ( 6(0.01t+3.6)^2 cm^3/cm )(0.01 cm/sec) = 0.06(0.01t+3.6)^2 cm^3/sec

= the rate of change of the volume per change in time, in cm^3/sec, after t seconds.


Problem to ponder: why is (V ∘ X)'(t) not a constant? Does the change in volume of a cube per change in side length depend on the side length?


Question part b)


Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.


This is a little ambiguous, but "use differentiation" suggests that we want y'(4.02) yunit per xunit, rather than Δy/Δx = (y(4.02)-y(4))/(0.02).


Neither of those make much sense, so I think we are to estimate Δy given x and Δx, without evaluating y(x) at all.

Then we want y'(x+Δx/2)×Δx


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

y'(x) = 4x + 3


y(4) = 44

y(4.02) = 44.3808

Δy = 0.3808

Δy/Δx = (0.3808)/(0.02) = 19.04


y'(4) = 19

y'(4.01) = 19.04

y'(4.02) = 19.08


Estimate Δy = (y(x+Δx)-y(x)/Δx without evaluating y() at all, using only y'(x), given x = 4, Δx = 0.02.


y'(x+Δx/2)×Δx = y'(4.01)×0.02 = 19.04×0.02 = 0.3808.


In this case, where y is quadratic in x, this method gives Δy exactly.

6 0
3 years ago
Quadrant:
NISA [10]

<h2>✒️Area Between Curves</h2>

\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}

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What are the new vertices of quadrilateral ABCD if the quadrilateral is reflected across the x-axis?
Alecsey [184]

Answer:

C.

Step-by-step explanation:

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Seventy percent of the engines manufactured on an assembly line are defective. Suppose engines are randomly selected one at a ti
strojnjashka [21]

Answer:

0.7

Step-by-step explanation:

Probability of defective = 0.7

P(non-defective) = 1 - 0.7 = 0.3

Let X be the trial at which we get the first non-defective

Then X follows a geometric distribution.

X ~ Geo (0.3)

What has happened already will not affect the probability.

P(atleast 2 more) = P(X > 1)

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