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3 years ago

Use the figure to find the measure of the numbered angles

Mathematics

1 answer:

ivanzaharov [21]3 years ago

5 0

Answer:

angle 2, 4, 6 are 34 degrees

angle 7, 5, 1, 3 are 146 degrees

180-34= 146

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HELP!<br>What is the distance between points (6, -9) and (-3, 4)?

ra1l [238]

Answer:

4 or $\sqrt{250}$

Step-by-step explanation:

I used M a t h w a y. it's really helpful. Good luck.

7 0

3 years ago

A swimming pool is being drained so it can be cleaned. The amount of water in the pool is changing according to the función f(t)

TiliK225 [7]

Answer:

0 ≤ t ≤ 5.

Step-by-step explanation:

In the function $f(t)$ , $t$ is the independent variable. The domain of $f$ is the set of all values of $t$ that this function can accept.

In this case, $f(t)$ is defined in a real-life context. Hence, consider the real-life constraints on the two variables. Both time and volume should be non-negative. In other words,

$t \ge 0$ .
$f(t) \ge 0$ .

The first condition is an inequality about $t$ , which is indeed the independent variable.

However, the second condition is about $f$ , the dependent variable of this function. It has to be rewritten as a condition about $t$ .

$\begin{aligned} f(t) &\ge 0 &&\text{Assumption} \cr 80000 - 16000\, t& \ge 0 && \text{Definition of} ~ f \cr 80000 & \ge 16000\, t && \begin{aligned}&\text{Add $16000\, t$} \\[-0.5em] & \text{to both sides of the inequality}\end{aligned} \cr 5 &\ge t &&\begin{aligned}&\text{Divide both sides of} \\[-0.5em] & \text{the inequality by $16000$}\end{aligned} \cr t &\le 5 && \text{Flip the inequality}\end{aligned}$ .

Hence, t ≤ 5.

Combine the two inequalities to obtain the domain:

0 ≤ t ≤ 5.

6 0

3 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$

8 0

3 years ago

Can you make a triangle with angles 45 degrees, 90 degrees, and 35 degrees?

nata0808 [166]

Answer:

Step-by-step explanation:

To make a triangle the angles have to add up to 180 degrees so 90+45+35=170 degrees not 180.

7 0

2 years ago

Choose the points that are solutions to this:

Scorpion4ik [409]

Answer:

b. (3,3)

Step-by-step explanation:

6 0

3 years ago