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3 years ago

6

hii!! can someone help me with this? pleasee is for TODAY it has to be a original one, i give brainliest too! Create a blueprint

of a home. but your priority is to provide appropriate measurements for each room. Your blueprint must be drawn to scale, and you have to provide a scale with or without units for it. You will use the blueprint and scale to determine the area of one room of your choice. This area will be displayed on your blueprint.

2 answers:

Softa [21]3 years ago

7 0

Answer:

wait how is that even possible i mean i will do it but how do i draw one.

Step-by-step explanation:

otez555 [7]3 years ago

4 0

Can I have a Scott sjfkfkf

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Find the difference quotient <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bf%28x%20%2B%20h%29%20-%20f%28x%29%7D%7Bh%7D" id="TexFor

AlladinOne [14]

Answer:

$-2x - h - 3$

Step-by-step explanation:

Step 1: Define

Difference Quotient: $\frac{f(x+h)-f(x)}{h}$

f(x) = -x² - 3x + 1

f(x + h) means that x = (x + h)

f(x) is just the normal function

Step 2: Find difference quotient

<u>Substitute:</u> $\frac{[-(x+h)^2-3(x+h)+1]-(-x^2-3x+1)}{h}$
<u>Expand and Distribute:</u> $\frac{[-(x^2+2hx+h^2)-3x-3h+1]+x^2+3x-1}{h}$
<u>Distribute:</u> $\frac{-x^2-2hx-h^2-3x-3h+1+x^2+3x-1}{h}$
<u>Combine like terms:</u> $\frac{-2hx-h^2-3h}{h}$
<u>Factor out </u><em><u>h</u></em><u>:</u> $\frac{h(-2x-h-3)}{h}$
<u>Simplify:</u> $-2x - h - 3$

8 0

3 years ago

Read 2 more answers

What does this equation equal to

Lena [83]

The answer to the equation is 115

6 0

3 years ago

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Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$

We use the initial condition $y(0)=-12$ to find the value of C.

$-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}$

Therefore,

$y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}$

4 0

3 years ago

What must be a factor of the polynomial function f(x) graphed on the coordinate plane below?

vesna_86 [32]

Answer:

x+2

Step-by-step explanation:

we know that

To find out the factors of a polynomial, determine the x-intercepts or zeros of the polynomial

Remember that

The x-intercepts of a polynomial are the values of x when the value of the polynomial is equal to zero

In this problem

Looking at the graph

For x=-2, f(x)=0

so

x=-2 is an x-intercept or zero of the polynomial

To find out the factor move the constant to the left side and equate to zero

$x=-2$

adds 2 both sides

$x+2=-2+2\\x+2=0$

therefore

The factor is (x+2)

5 0

3 years ago

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Item 1<br> Round 71.58043 to the nearest hundredth

Natalija [7]

Answer:

71.58

Step-by-step explanation:

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3 years ago