-4 times a number plus 16 is -116. find the number and show work.

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Licemer1 [7]

Answer:

222.7 cm³ of paper to wrap this.

Step-by-step explanation:

The rectangular prism is SA=2(wl+hl+hw)

Where:

w=5 l=5 h=4

SA= 2((5*5)+(4*5)+(4*5))

SA=2(25+20+20)

SA=2(65)=130cm³

The square base pyramid is $A= lw+l\sqrt{x} (w/2)^2+h^2+w +w\sqrt{x} (1/w)^2+h^2$

$SA= 6*5+6*\sqrt{x} (5/2)^2+5^2+5*\sqrt{x} (6/2)^2$

≈92.69578 cm³

Add the two together 130+92.70 =222.7 cm³

5 0

4 years ago

Read 2 more answers

An English class attended a performance at the theater. The class took 25 students and 3 adults. Each student ticket, s, cost $3

castortr0y [4]

Total cost = student cost + adult cost
t = 25s + 3(s+3)
t=28s+9

8 0

4 years ago

750x + 17y = 141.61

MAVERICK [17]

I’ll answer this in a second

7 0

3 years ago

Read 2 more answers

What expression is quadratic monomial?

grandymaker [24]

Answer:

the solution is x^4. the second one

7 0

3 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

3 0

3 years ago