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3 years ago

PLZZZ HELPP WILL REWARD BRAINLEST

Mathematics

1 answer:

Kipish [7]3 years ago

3 0

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(Ⅰ) 2v² - 2v³ - 7 + 7v :

Standard Form → 2v³ + 2v² + 7v - 7v

Number of terms → Four (Polynomial)

Degree of Polynomial → Highest degree 3 (Cubic)

(Ⅱ) -3m -8m -7m²

Standard Form → -7m² - 8m - 3m

Number of terms → Three (Trinomial)

Degre of Polynomial → Highest degree 2 (Quadratic)

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Can someone help me with this thanks

Ugo [173]

The length of the brace required is 4.3m

What is sine rule?

In a ΔABC a, b and c are the sides and A, B and C are angles then,

$\frac{a}{SinA}=\frac{b}{sinB}=\frac{c}{sinC}$

We can find the length, l as shown below:

Let AB=3m, BC=2m and AC=l

Let ∠A=25°

So, in ΔABC

$\frac{BC}{sinA}=\frac{AB}{sinC}=\frac{AC}{SinB}$

$\frac{BC}{sinA}=\frac{AB}{sinC}$

$\frac{2}{sin25}=\frac{3}{sinC}$

$\angle{C}=sin^{-1}(\frac{3\times sin25}{2} )$

∠C=39.34°

∠A+∠B+∠C=180°

∠B=180°-25°-39.34°

∠B=115.66°

$\frac{BC}{sinA}=\frac{AC}{SinB}$

$\frac{2}{sin25}=\frac{l}{sin(115.66)}$

$l=\frac{2\times sin(115.66)}{sin25}$

l=4.2659

Rounding to nearest tenth of meter.

l=4.3m

Hence, the length of the brace required is 4.3m

Learn more about Sine Rule here:

brainly.com/question/25852087

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8 0

2 years ago

Taylor surveys students in one grade level who own at least one pet. She finds that 50% of the students surveyed own 2 pets, 3 s

Llana [10]

First, we need to work out the total number of students who were being surveyed.

We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population

That means total number of students being surveyed is 13+13=26 students

Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13

The probability distribution is shown in the table below. Let $x$ be the number of pets and $P(X=x)$ is the probability of owning the number of pets