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3 years ago

Please help ive been stuck!!

Mathematics

1 answer:

Kobotan [32]3 years ago

8 0

I would say no, if you follow the lines on the graph, 12 on the x axis and 5 on the Y axis they are not within the range indicated in gray.

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Work out the area of this triangle. I will mark you as brainliest, please help quick.

nataly862011 [7]

Answer:

22 cm²

Step-by-step explanation:

Work out the area of this triangle.

Area Formula: 1/2b * h where b = base (11cm) and h = height (4 cm)

1/2 *11 * 4 =

5.5 * 4 =

22 cm²

3 0

1 year ago

21. Estimate the annual wages for a cashier who earns $11 an hour.

DerKrebs [107]

Answer:

~22,000 /yr

Step-by-step explanation:

52 weeks/yr

5 day work week (8 hrs a day 40hrs a week)

comes out to 22,880 but mandatory holidays would be lower

3 0

2 years ago

two angles are supplementary the first angle measures 40 degree what's the measurement of the second angle

Leni [432]

Supplementary is when 2 angles add up to 180°.
180°-40°=140°

4 0

2 years ago

Read 2 more answers

A small box weighs 1 25 pounds. A large box weighs 3 1/3 time as much as the small box. how much does the large box weigh?

egoroff_w [7]

Answer:

416.7pounds

Step-by-step explanation:

Given parameters:

Weight of the small box = 125pounds

Unknown:

Weight of the larger box = ?

Solution:

The large box weighs $3\frac{1}{3}$ times as the small box:

So;

= 125pounds x $\frac{10}{3}$

= 416.7pounds

6 0

2 years ago

Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .

Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of $F$ around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use the Green's Theorem to evaluate this integral. The rectangle is presented below.

We have that

$F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle$

Therefore,

$P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y$

Let's calculate the needed partial derivatives.

$P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0$

Thus,

$Q_x -P_y = 0 -2 = - 2$

Now, by the Green's theorem, we have

$\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x \Big|_{-3}^{4} = -42$

4 0

2 years ago