All categories

3 years ago

Enter the giveaway for points

Mathematics

2 answers:

Nookie1986 [14]3 years ago

8 0

Answer:

Thxx for points how do I enter

Step-by-step explanation:

Nikolay [14]3 years ago

4 0

Answer:

OK! :DDD

Step-by-step explanation:

You might be interested in

Which image below is an acute triangle?

lawyer [7]

I don’t see any images but any angle measured less than 90 degrees is an acute angle

4 0

3 years ago

The equation P = 31 + 1.75w models the relation between the amount of Tuyet's monthly water bill payment, P, in

dimulka [17.4K]

Answer:

The "P" intercept, 31, means that Tuyet's monthly water bill would be $31 if he used no water.

Step-by-step explanation:

5 0

2 years ago

Alice works at a health care facility. She has measured the hemoglobin levels of 200 people. The data follows a normal distribut

dimulka [17.4K]

In normal distribution, 68% of the values are within +_ 1 SD. That leaves 32% outside that range. Thus, 32 people with greater than 13, 100 people greater than 14 (200*0.68/2+32)

5 0

2 years ago

Read 2 more answers

Jayden buys a game for $10.77 and some lotion for $3.12. He pays with a $20 bill. Estimate how much change he should get, to

guapka [62]

Answer:

7.65

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Question 15 (5 points)<br> Find the angle between u = <7, -2> and v= (-1,2>.

Tema [17]

Answer:

Approximately $2.3127$ radians, which is approximately $132.51^{\circ}$ .

Step-by-step explanation:

Dot product between the two vectors:

$\begin{aligned}& u\cdot v \\ =\; & 7 \times (-1) + (-2) \times 2 \\ =\; & (-11) \end{aligned}$ .

Magnitude of the two vectors:

$\begin{aligned} \| u \| &= \sqrt{{7}^{2} + {(-2)}^{2}} \\ &= \sqrt{53} \end{aligned}$ .

$\begin{aligned} \| v \| &= \sqrt{{(-1)}^{2} + {2}^{2}} \\ &= \sqrt{5} \end{aligned}$ .

Let $\theta$ denote the angle between these two vectors. By the property of dot products:

$\begin{aligned} \cos(\theta) &= \frac{u \cdot v}{\|u\| \, \| v \|} \\ &= \frac{(-11)}{(\sqrt{53})\, (\sqrt{5})} \\ &= \frac{(-11)}{\sqrt{265}}\end{aligned}$ .

Apply the inverse cosine function ${\rm arccos}$ to find the value of this angle:

$\begin{aligned} \theta &= \arccos\left(\frac{u \cdot v}{\| u \| \, \| v \|}\right) \\ &= \arccos\left(\frac{(-11)}{\sqrt{265}}\right) \\ & \approx \text{$2.3127$ radians} \\ &= 2.3127 \times \frac{180^{\circ}}{\pi} \\ &\approx 132.51^{\circ}\end{aligned}$ .

8 0

2 years ago