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3 years ago

Can someone that is good at 7th grade math help me plzzzz

Mathematics

1 answer:

stiks02 [169]3 years ago

6 0

Answer:

-4 *-4 *-4

= 16*-4

= ANWSER

-9*-9

=81

Step-by-step explanation:

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Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,

Anton [14]

Answer:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

Step-by-step explanation:

So we have the equation:

$\tan(x-y)=\frac{y}{8+x^2}$

And we want to find dy/dx.

So, let's take the derivative of both sides:

$\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[\tan(x-y)]$

We can use the chain rule, where:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

$=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])$

Differentiate x like normally. Implicitly differentiate for y. This yields:

$=\sec^2(x-y)(1-y')$

Distribute:

$=\sec^2(x-y)-y'\sec^2(x-y)$

And that is our left side.

Right Side:

We have:

$\frac{d}{dx}[\frac{y}{8+x^2}]$

We can use the quotient rule, where:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

f is y. g is (8+x²). So:

$=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}$

Differentiate:

$=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

And that is our right side.

So, our entire equation is:

$\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

$((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)$

The right side cancels. Let's distribute the left:

$\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy$

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

$\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2$

Move -2xy to the left. So:

$\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2$

Factor out a y' from the right:

$\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)$

Divide. Therefore, dy/dx is:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}$

We can factor out a (8+x²) from the denominator. So:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

And we're done!

8 0

3 years ago

Proportions in Triangles

alukav5142 [94]

Answer:

x =35.

Step-by-step explanation:

x / 20 = 14 / 8

Cross multiply:

8x = 20*14

8x = 280

x = 35.

3 0

3 years ago

x, y, a, and b are positive integers. When x is divided by y, the remainder is 6. When a is divided by b, the remainder is 9. Wh

Mademuasel [1]

Answer:

y + b > 15

Step-by-step explanation:

given data

positive integers = x, y, a, b

x divided by y then remainder = 6

a divided by b then remainder = 9

to find out

possible value for y + b

solution

we know that here when x divided by y then remainder is 6

its mean y is greater than 6 and when a divided by b then remainder is 9

so its mean b is greater than 9

and we know remainder is less than divisor

so here y + b must be greater than = 6 + 9

y + b > 15

5 0

3 years ago

Find the product <br> -1/2 y(2y3-8)

MissTica

$0-(( \frac{1}{2} *y)*2y^3-8) \\ \\ then \ we \ would \ simplify \ 1/2 \\ \\ we \ factor y^3 - 4 \\ \\ 4 \ would \ not \ be \ a \ perfect \ cube. \\ \\ therefore, \ your \ answer \ would \ then \ be \ the \ following: \\ \\ \boxed{ -y * (y^3 - 4)}$

8 0

3 years ago

If sin 21 = cos (5x+9) what is the value of x

n200080 [17]

sin x = cos 54 = sin (90-54) = sin 46

Hence x = 46 deg

Same technique. Sin 21 = cos (5x+9) = sin (90-5x-9)

81-5x = 21

Or 5x = 60

Or x = 12 deg

hope it helps

4 0

3 years ago