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2 years ago

6

Pls Help This is Due Soon (4/27): The following ordered pairs from a linear function: (-5,-7), (-2, 2), (3, 17), (7, 29), аnd (1

2, 44). What is the y-intercept of the
function? *
-7
7
0 8

2 answers:

erastovalidia [21]2 years ago

6 0

Answer:

8 because this is a linear function meaning that it is being added by a certain number. If you find the slope using the formula x2-x1/ y1-y2 you would get 3. If you add 3 for every term starting from -5 until you get to 0.

lesantik [10]2 years ago

4 0

-7 ur welcome

can you please mark me as brainlist

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Can someone help me with this question

Mashcka [7]

The correct answer here would be D. -1(3x+1)(x+5). You can find this answer by redistributing the problem.

-1(3x+1)(x+5)
-1(3x²+15x+1x+5)
-1(3x²+16x+5)
-3x²-16x-5

Using the math above, we can see that when we redistribute we get the original equation. That makes Choice D correct.

5 0

3 years ago

What is all of the surface area and volume of this Castle? Find the surface area and volume of all the figures below, then out o

motikmotik

Answer:

Step-by-step explanation:

There are a few formulas that are useful for this:

lateral area of a pyramid or cone: LA = 1/2·Ph, where P is the perimeter and h is the slant height
lateral area of a cylinder: LA = π·dh, where d is the diameter and h is the height
area of a rectangle: A = lw, where l is the length and w is the width
volume of a cone or pyramid: V = 1/3·Bh, where B is the area of the base and h is the height
volume of a cylinder or prism: V = Bh, where B is the area of the base and h is the height

You will notice that for lateral area purposes, a pyramid or cone is equivalent to a prism or cylinder of height equal to half the slant height. And for volume purposes, the volume of a pyramid or cone is equal to the volume of a prism or cylinder with the same base area and 1/3 the height.

Since the measurements are given in cm, we will use cm for linear dimensions, cm^2 for area, and cm^3 for volume.

___

The heights of the cones at the top of the towers can be found from the Pythagorean theorem.

(slant height)^2 = (height)^2 + (radius)^2

height = √((slant height)^2 - (radius)^2) = √(10^2 -5^2) = √75 = 5√3

The heights of the pyramids can be found the same way.

height = √(13^2 -2^2) = √165

___

<u>Area</u>

The total area of the castle will be ...

total castle area = castle lateral area + castle base area

These pieces of the total area are made up of sums of their own:

castle lateral area = cone lateral area + pyramid lateral area + cylinder lateral area + cutout prism lateral area

and ...

castle base area = cylinder base area + cutout prism base area

So, the pieces of area we need to find are ...

cone lateral area (2 identical cones)
pyramid lateral area (2 identical pyramids)
cylinder lateral area (3 cylinders, of which 2 are the same)
cutout prism lateral area
cylinder base area (3 cylinders of which 2 are the same)
cutout prism base area

Here we go ...

Based on the above discussion, we can add 1/2 the slant height of the cone to the height of the cylinder and figure the lateral area of both at once:

area of one cone and cylinder = π·10·(18 +10/2) = 230π

area of cylinder with no cone = top area + lateral area = π·1^2 +π·2·16 = 33π

area of one pyramid = 4·4·(13/2) = 52

The cutout prism outside face area is equivalent to the product of its base perimeter and its height, less the area of the rectangular cutouts at the top of the front and back, plus the area of the inside faces (both vertical and horizontal).

outside face area = 2((23+4)·11 -3·(23-8)) = 2(297 -45) = 504

inside face area = (3 +(23-8) +3)·4 = 84

So the lateral area of the castle is ...

castle lateral area = 2(230π + 52) +33π + 504 + 84 = 493π +692

≈ 2240.805 . . . . cm^2

The castle base area is the area of the 23×4 rectangle plus the areas of the three cylinder bases:

cylinder base area = 2(π·5^2) + π·1^2 = 51π

prism base area = 23·4 = 92

castle base area = 51π + 92 ≈ 252.221 . . . . cm^2

Total castle area = (2240.805 +252.221) cm^2 ≈ 2493.0 cm^2

___

<u>Volume</u>

The total castle volume will be ...

total castle volume = castle cylinder volume + castle cone volume + castle pyramid volume + cutout prism volume

As we discussed above, we can combine the cone and cylinder volumes by using 1/3 the height of the cone.

volume of one castle cylinder and cone = π(5^2)(18 + (5√3)/3)

= 450π +125π/√3 ≈ 1640.442 . . . . cm^3

volume of flat-top cylinder = π·1^2·16 = 16π ≈ 50.265 . . . . cm^3

The volume of one pyramid is ...

(1/2)4^2·√165 = 8√165 ≈ 102.762 . . . . cm^3

The volume of the entire (non-cut-out) castle prism is the product of its base area and height:

non-cutout prism volume = (23·4)·11 = 1012 . . . . cm^3

The volume of the cutout is similarly the product of its dimensions:

cutout volume = (23 -8)·4·3 = 180 . . . . cm^3

so, the volume of the cutout prism is ...

cutout prism volume = non-cutout prism volume - cutout volume

= 1012 -180 = 832 . . . . cm^3

Then the total castle volume is ...

total castle volume = 2·(volume of one cylinder and cone) + (volume of flat-top cylinder) +2·(volume of one pyramid) +(cutout prism volume)

= 2(1640.442) + 50.265 +2(102.762) +832 ≈ 4368.7 . . . . cm^3

4 0

3 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

3 years ago

The volume of a sphere is two-thirds of a cylinder. What is the volume of a cylinder that has the same diameter as a sphere with

sattari [20]

Answer:

$V_2= \frac{28}{3} u^3$

Step-by-step explanation:

Given

Represent the volume of the cylinder with V1 and the volume of the sphere with V2

So, from the first statement: we have:

$V_2 =\frac{2}{3}V_1$

and

$V_1 = 14u^3$

To solve for $V_2$ , we simply substitute $14u^3$ for $V_1$ in $V_2 =\frac{2}{3}V_1$

$V_2 =\frac{2}{3}V_1$

$V_2= \frac{2}{3} * 14u^3$

$V_2= \frac{2* 14}{3} u^3$

$V_2= \frac{28}{3} u^3$

Hence, the volume of the sphere is $\frac{28}{3} u^3$

5 0

2 years ago

Find the least common multiple of (5x+15) and (3x+9)

JulsSmile [24]

X and 3 is the answer i think

5 0

3 years ago

Read 2 more answers