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3 years ago

9

Barry is given $0.40 and $0.90 from his parents. he adds this to the $6.30 in his piggy bank how much money does he have in toto

al

1 answer:

xeze [42]3 years ago

8 0

Answer:

$7.30

Step-by-step explanation:

If you add .40 + .60 you'll get 1 and if you add that to the 6.30 then you'll get $7.30

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Please answer this question now in two minutes

fenix001 [56]

Answer:

Equation of a line is y = mx + c

where

m = slope

c = y intercept

y = 1/6x - 3

Comparing with the above formula

m = 1/6

Since the lines are parallel their slope are also the same

So the equation of the line using point (-4,2) is

y - 2 = 1/6( x + 4)

y - 2 = 1/6x + 2/3

y = 1/6x + 2/3 + 2

y = 1/6x + 8/3

Hope this helps

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3 years ago

BRAINLIESTTT ASAP! PLEASE HELP ME :)

11Alexandr11 [23.1K]

<h3>Answer: B) Only the first equation is an identity</h3>

========================

I'm using x in place of theta. For each equation, I'm only altering the left hand side.

Part 1

cos(270+x) = sin(x)

cos(270)cos(x) - sin(270)sin(x) = sin(x)

0*cos(x) - (-1)*sin(x) = sin(x)

0 + sin(x) = sin(x)

sin(x) = sin(x) ... equation is true

Identity is confirmed

---------------------------------

Part 2

sin(270+x) = -sin(x)

sin(270)cos(x) + cos(270)sin(x) = -sin(x)

-1*cos(x) + 0*sin(x) = -sin(x)

-cos(x) = -sin(x)

We don't have an identity. If the right hand side was -cos(x), instead of -sin(x), then we would have an identity.

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3 years ago

Help ill give brainliest!!!!!!!<br><br> ^^

Strike441 [17]

Answer:

15 days

Step-by-step explanation:

Well, as you would be able to tell, you just have to find the LCM of 3 and 5. These do not have one before 3 x 5, so it is 15.

3 0

3 years ago

The pressure applied to a leverage bar varies inversely as the distance from the object. If 150 pounds is required for a distanc

AlexFokin [52]

Answer:

500 pounds

Step-by-step explanation:

Let the pressure applied to the leverage bar be represented by p

Let the distance from the object be represented by d.

The pressure applied to a leverage bar varies inversely as the distance from the object.

Written mathematically, we have:

$p \propto \dfrac{1}{d}$

Introducing the constant of proportionality

$p = \dfrac{k}{d}$

If 150 pounds is required for a distance of 10 inches from the object

p=150 pounds
d=10 inches

$150 = \dfrac{k}{10}\\\\k=1500$

Therefore, the relationship between p and d is:

$p = \dfrac{1500}{d}$

When d=3 Inches

$p = \dfrac{1500}{3}\\\implies p=500$ pounds$

The pressure applied when the distance is 3 inches is 500 pounds.

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3 years ago

Find the lateral area for the pyramid with the equilateral base

likoan [24]

<h3>The lateral area for the pyramid with the equilateral base is 144 square units</h3>

<em><u>Solution:</u></em>

The given pyramid has 3 lateral triangular side

The figure is attached below

Base of triangle = 12 unit

<em><u>Find the perpendicular</u></em>

By Pythagoras theorem

$hypotenuse^2 = opposite^2 + adjacent^2$

Therefore,

$opposite^2 = 10^2 - 6^2\\\\opposite^2 = 100 - 36\\\\opposite^2 = 64\\\\opposite = 8$

<em><u>Find the lateral surface area of 1 triangle</u></em>

$\text{ Area of 1 lateral triangle } = \frac{1}{2} \times opposite \times base$

$\text{ Area of 1 lateral triangle } = \frac{1}{2} \times 8 \times 12\\\\\text{ Area of 1 lateral triangle } = 48$

<em><u>Thus, lateral surface area of 3 triangle is:</u></em>

3 x 48 = 144

Thus lateral area for the pyramid with the equilateral base is 144 square units

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3 years ago