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2 years ago

Please 100 points and brainliest if u do them all plsssssss I rlly need help asap

Mathematics

1 answer:

AysviL [449]2 years ago

7 0

Answer:

<u>Question 1</u>

The volume of figure A can be found by multiplying (5 · 7)(2).

<u>Question 2</u>

The total volume of the figure is 442 mm³

<u>Question 3</u>

The volume of the triangular prism is 108 in³

Step-by-step explanation:

<u>Formula used</u>

Volume of a prism = base area × height

Area of a rectangle = width × length

Area of a triangle = 1/2 × base × height

Volume of a cube = s³ (where s is the side length)

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

<h3><u>Question 1</u></h3>

⇒ Volume of Figure A = (5 · 7)(2)

= 70 cm³

⇒ Volume of Figure B = (13 · 7)(2)

= 182 cm³

⇒ Total Volume = Volume of Figure A + Volume of Figure B

= 70 + 182

= 252 cm³

The first statement is false.

<u>Rewritten statement</u>:

The volume of figure A can be found by multiplying (5 · 7)(2).

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

<h3><u>Question 2</u></h3>

⇒ Volume of rectangular prism = (15 · 15)(2)

= 450 mm³

⇒ Volume of central cube = 2³

= (2 · 2 · 2)

= 8 mm³

⇒ Total Volume = Volume of rectangular prism - Volume of cube

= 450 - 8

= 442 mm³

The third statement is false.

<u>Rewritten statement</u>:

The total volume of the figure is 442 mm³

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

<h3><u>Question 3</u></h3>

⇒ Volume of rectangular prism = (12 · 3)(8)

= 288 in³

⇒ Volume of triangular prism = (1/2 · 12 · 6)(3)

= 108 in³

⇒ Total Volume = Volume of rectangular prism + Volume of triangular prism

= 288 + 108

= 396 in³

The second statement is false.

<u>Rewritten statement</u>:

The volume of the triangular prism is 108 in³

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notsponge [240]

Complete Question

ymposium is part of a larger work referred to as Plato's Dialogues. Wishart and Leach† found that about 21.4% of five-syllable sequences in Symposium are of the type in which four are short and one is long. Suppose an antiquities store in Athens has a very old manuscript that the owner claims is part of Plato's Dialogues. A random sample of 498 five-syllable sequences from this manuscript showed that 129 were of the type four short and one long. Do the data indicate that the population proportion of this type of five-syllable sequence is higher than that found in Plato's Symposium? Use = 0.01.

a. What is the value of the sample test statistic? (Round your answer to two decimal places.)

b. Find the P-value of the test statistic. (Round your answer to four decimal places.)

Answer:

a) $Z=2.45$

b) $P Value=0.0073$

Step-by-step explanation:

From the question we are told that:

Probability of Wishart and Leach $P=21.4=>0.214$

Population Size $N=498$

Sample size $n=12$

Therefore

$P'=\frac{129}{498}$

$P'=0.2590$

Generally the Null and Alternative Hypothesis is mathematically given by

$H_0:P=0.214$

$H_a:=P>0.214$

Test Statistics

$Z=\frac{P'-P}{\sqrt{\frac{P(1-P)}{n}}}$

$Z=\frac{0.2590-0.214}{\sqrt{\frac{0.214(1-0.214)}{498}}}$

$Z=2.45$

Therefore P Value is given as

$P Value =P(Z\geq 2.45)$

$P Value =1-P(Z\leq 2.45)$

$P Value =1-0.99268525$

$P Value=0.0073$

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3 years ago

While packing for thier cross-country move, the chen family uses a crate that has the shape of a cube. The Chens want to pack a

Brrunno [24]

If the framed picture is shaped like a square and has a 12 square foot surface area, then the answer is yes, it will fit flush against the edge of the crate.

Given Part A:

the volume of the cube = 64 cubic feet

therefore, ∛64 = 4 feet

hence one edge measures 4 feet.

Now for Part B:

the area of the square is 12 square feet.

hence, √12 = 3.36 feet.

we can observe that 3.46<4

which indicates that the area covered by the painting is less than that of the one side of the crate, which makes it easy for the painting to fit in the crate.

Hence the painting will fit a side of crate.

Learn more about Area and Perimeter here:

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While packing for their cross-country move, the Chen family uses a that has the shape of a cube. PART A PART B If the crate has the volume V = 64 cubic feet, The Chens want to pack a large, framed painting. If an area of 12 square feet , will the painting fit flat what is the length of one edge? the framed painting has the shape of a square with against a side of the crate? Explain.

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2 years ago

1. The height of a triangle is 6 m more than its base. The area of the triangle is 56 m². What is the length of the base? Enter

Elodia [21]

Answers:
1. 8 m
2. 17 m
3. 7 cm
4. 2 s

Explanations:

1. Let x = length of the base
x + 6 = height of the base

Then, the area of the triangle is given by

(Area) = (1/2)(base)(height)
56 = (1/2)(x)(x + 6)
56 = (1/2)(x²  + 6x)

Using the symmetric property of equations, we can interchange both sides of equations so that

(1/2)(x²  + 6x) = 56

Multiplying both sides by 2, we have

x² + 6x = 112

The right side should be 0. So, by subtracting both sides by 112, we have

x² + 6x - 112 = 112 - 112
x² + 6x - 112 = 0

By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation becomes

(x - 8)(x +14) = 0

So, either

x - 8 = 0 or x + 14 = 0

Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.

2. Let x = length of increase in both length and width of the rectangular garden

Then,

14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden

So,

(Area of the new garden) = (length of the new garden)(width of the new garden)

255 = (14 + x)(12 + x) (1)

Note that

(14 + x)(12 + x) = (x + 14)(x + 12)
= x(x + 14) + 12(x + 14)
= x² + 14x + 12x + 168
= x² + 26x + 168

So, the equation (1) becomes

255 = x² + 26x + 168

By symmetric property of equations, we can interchange the side of the previous equation so that

x² + 26x + 168 = 255

To make the right side becomes 0, we subtract both sides by 255:

x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0

To solve the preceding equation, we use the quadratic formula.

First, we let

a = numerical coefficient of x² = 1

Note: if the numerical coefficient is hidden, it is automatically = 1.

b = numerical coefficient of x = 26
c = constant term = - 87

Then, using the quadratic formula

$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-26 \pm \sqrt{26^2 - 4(1)(-87)} }{2(1)} 
\newline x = \frac{-26 \pm \sqrt{1,024} }{2}
\newline
\newline x = \frac{-26 \pm 32 }{2}$

So,

$x = \frac{-26 + 32 }{2} \text{ or } x = \frac{-26 - 32 }{2}
\newline x = \frac{6 }{2} \text{ or } x = \frac{-58 }{2}
\newline \boxed{ x = 3 \text{ or } x = -29}$

Since x represents the amount of increase, x should be positive.

Hence x = 3.

Therefore, the length of the new garden is given by

14 + x = 14 + 3 = 17 m.

3. The area of the shaded region is given by

(Area of shaded region) = π(outer radius)² - π(inner radius)²
= π(2x)² - π6²
= π(4x² - 36)

Since the area of the shaded region is 160π square centimeters,

π(4x² - 36) = 160π

Dividing both sides by π, we have

4x² - 36 = 160

Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.

Adding both sides of the equation by 36, we have

4x² - 36 + 36 = 160 + 36
4x² = 196

Then, we divide both sides by 4 so that

x² = 49

Taking the square root of both sides, we have

$x = \pm 7$

Note: If we take the square root of both sides, we need to add the plus minus sign $(\pm)$ because equations involving x² always have 2 solutions.

So, x = 7 or x = -7.

But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.

Hence x = 7 cm

4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,

h(t) = 0 (1)

Moreover, since $v_0 = 27$ and $h_0 = 10$ ,

h(t) = -16t² + 27t + 10 (2)

Since the right side of the equations (1) and (2) are both equal to h(t), we can have

-16t² + 27t + 10 = 0

To solve this equation, we'll use the quadratic formula.

Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula.

First, we let

a = numerical coefficient of t² = -16
b = numerical coefficient of t = 27
c = constant term = 10

Then, using the quadratic formula

$t = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-27 \pm \sqrt{27^2 - 4(-16)(10)} }{2(-16)} \newline t = \frac{-27 \pm \sqrt{1,369} }{-32} \newline \newline t = \frac{-27 \pm 37 }{32}$

So,

$t = \frac{-27 + 37 }{-32} \text{ or } t = \frac{-27 - 37 }{-32} \newline t = \frac{-10}{32} \text{ or } t = \frac{-64 }{-32} \newline \boxed{ t = -0.3125 \text{ or } t = 2}$

Since t represents the amount of time, t should be positive.

Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.

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3 years ago

Read 2 more answers

The distance from the Earth to the Sun is approximately 90,000,000 miles, which can be written as 9 × 10a miles, where a = . The

andrey2020 [161]

10*a = 10000000

a = 1000000, one million

5*10*b = 500000000

b = 10000000, ten million

About ten times.

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3 years ago