All categories

3 years ago

Help please its timed

Mathematics

1 answer:

igomit [66]3 years ago

7 0

Answer:

Step-by-step explanation:

Order of Operations: BPEMDAS

Step 1: Write out expression

2² + 6(3) - 7

Step 2: Evaluate exponents

4 + 6(3) - 7

Step 3: Evaluate parenthesis (multiply)

4 + 18 - 7

Step 4: Add

22 - 7

Step 5: Subtract

You might be interested in

Factor completely 12x^2 - 9x + 15

alexgriva [62]

Answer:

3(4x^2-3x+5)

Step-by-step explanation:

3(4x^2-3x+5)

8 0

3 years ago

Read 2 more answers

A rational function has x = - 4 as a vertical asymptote. What term appears in factored form in the denominator of the function?

pantera1 [17]

Answer:

A.x+4

Step-by-step explanation:

We are given that a rational function which has a vertical asymptote at x=-4.

We have to find the term which appears in factored form in the denominator of the function.

We know that

To find the vertical asymptote of a rational function we will substitute the

denominator equal to zero.

Then, solve for x.

The value of x obtained are called vertical asymptotes.

By definition of vertical asymptote

x=-4

x+4=0

x+4 will be appears in factored form in the denominator of the function.

5 0

4 years ago

I need help with 3 and 4 pls

vladimir2022 [97]

Answer U cant answer 4 without answering 3 good luck

Step-by-step explanation:

7 0

3 years ago

Determine the values of the unlabeled parts of the triangle.

____ [38]

Answer:

∠Q = 63.4º, PQ = 2, QO = 4.47

Step-by-step explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as the equilateral triangle, isosceles triangle, scalene triangle, right angled triangle.

Triangle POQ is a right angled triangle because ∠P = 90°. Also, ∠O = 26.6°.

Therefore in ΔPOQ: ∠P + ∠O + ∠Q = 180° (sum of angles in a triangle).

90 + 26.6 + ∠Q = 180

∠Q + 116.6 = 180

∠Q = 180 - 116

∠Q = 63.4°

We can find PQ and QO using sine rule. Sine rule for ΔPOQ is given as:

$\frac{PO}{sin Q}= \frac{PQ}{sinO}= \frac{QO}{sin P}\\\\Finding\ PQ :\\\\\frac{PO}{sin Q}= \frac{PQ}{sinO}\\\\\frac{4}{sin(63.4)} =\frac{PQ}{sin(26.6)}\\\\PQ= \frac{4}{sin(63.4)} *sin(26.6)\\\\PQ = 2\\\\Also\ finding\ QO:\\\\\frac{PO}{sin Q}= \frac{QO}{sinP}\\\\\frac{4}{sin(63.4)} =\frac{QO}{sin(90)}\\\\QO=\frac{4}{sin(63.4)} *sin(90)\\\\QO = 4.47$