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3 years ago

Pls help I’ll brainlest grades close today pls helpp

Mathematics

1 answer:

kow [346]3 years ago

3 0

Answer:

First deal is better, unit rate of first deal is 22

Step-by-step explanation:

176/8=22

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A client would like this logo printed onto a canvas that is at least 70 inches tall. The original logo is 4.5 inches wide by 3.6

vfiekz [6]

Answer:

87.5 by 70 inches

Step-by-step explanation:

No options were given. So, I will calculate the minimum width

Given

$Original\ Logo = 4.5 : 3.6$

Height = 70 in ---- of the printed logo

Required

Determine the dimension that keeps the requirement

Let x be the width of the printed logo.

So, the ratio can be represented as:

$Printed\ Logo = x : 70$

Equate both ratios

$x : 70 = 4.5 : 3.6$

As fraction

$x / 70 = 4.5 / 3.6$

Multiply through by 70

$70 * x / 70 = 4.5 / 3.6* 70$

$x = 87.5$

So, one of the dimension that meets the requirement is a width of 87.5 inches

4 0

3 years ago

Can anyone help me on this!

icang [17]

it should be B

if i’m wrong please correct me

4 0

4 years ago

Space shuttle Challengerexploded because of O-ring failure shortly after it was launched. O-ring damage and temperature at time

arlik [135]

Answer:

The mean launch temperature for flights with O-ring damage is significantly lesser than Flights with No O-ring

Step-by-step explanation:

Let

Null hypothesis : H0 : μ1 ≥ μ2

Alternate hypothesis : Ha : μ1 < μ2

next calculate the test statistic

t = -2.312

calculate the P-value

p-value = T.DIST ( t ,df ) ( excel function )

= 0.0270

Given that P-value ( 0.020 ) < level of significance ( 0.05 ) we will not fail to reject the null hypothesis

Attached below is the detailed solution

3 0

3 years ago

Use a coterminal angle to find the exact value of the expression. Do not use a calculator.

melamori03 [73]

<span>c.
d.</span><span>Seleccione una respuesta.</span>

5 0

3 years ago

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago