1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ratling [72]
3 years ago
12

What is the verbal expression for this problem?

Mathematics
2 answers:
hodyreva [135]3 years ago
5 0

Answer:

3(9+r)

Step-by-step explanation:

pretty sure that should be it if not then mb

maks197457 [2]3 years ago
5 0
3(9+r) hope that helps .

Explanation: since it says times and then the sum of you know you need to put parenthesis and that the 3 would go on the outside, and you know you need to add
You might be interested in
A 24-foot utility pole casts a 32-foot shadow. To the nearest degree, find the angle of elevation of the sun. A) 37° B) 41° C) 4
Lelechka [254]
The answer is 49 degrees, so C.
5 0
3 years ago
Read 2 more answers
Add. 5 1/8 + 5/6 + 5/12. <br><br>A) 6 3/4<br><br>B) 6 3/8<br><br>C) 6 3/12<br><br>D) 6 3/24
djverab [1.8K]
Let’s change all of the denominators to be 48. He now have 6/48+40/48+20/48, or 66/48. This becomes 33/24, and in total 5+33/24, or 6 and 9/24. 9/24 becomes 3/8, so your answer is B.
5 0
4 years ago
How would I convert this function into an equation ?
vivado [14]

Definition

Functions have zeros or x-values that generate function values equal to zero.

Equations have roots or solutions that can be substituted into the polynomial in order to generate a value equal to zero.

Given:

f(x)\text{ = -}\frac{1}{3}\sqrt[]{x}\text{ + 3}

To convert this to an equation, we set f(x) equal to zero i.e

f(x)\text{ = 0}

The resulting equation is:

-\text{ }\frac{1}{3}\sqrt[]{x}\text{   + 3 = 0}

5 0
1 year ago
A student claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. Explain the student's m
____ [38]

Answer:

The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i

Step-by-step explanation:

1) This claim is mistaken.

2) The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.

a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}

For example:

3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

\Delta < 0\\b^{2}-4*a*c

We'll have <em>n </em>different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

\Delta < 0

-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)

2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\

4 0
3 years ago
Write the vector u as a sum of two orthogonal vectors, one of which is the vector projection of u onto v, proj,u
11Alexandr11 [23.1K]

Answer:

Take b = \frac{-17}{25}(7,1) as the projection of u onto v and w \frac{1}{25}(-31,217) as the vector such that b+w =u

Step-by-step explanation:

The formula of projection  of a vector u onto a vector v is given by

\frac{u\cdot v}{v\cdot v}v, where cdot is the dot product between vectors.

First, let b ve the projection of u onto v. Then

b = \frac{u\cdot v}{v\cdot v}v= \frac{-6\cdot 7+8\cdot 1}{7\cdot 7+1\cdot 1}(7,1) = \frac{-34}{50}(7,1) = \frac{-17}{25}(7,1)

We want a vector w, that is orthogonal to b and that b+w = u. From this equation we have that w = u-b = (-6,8)-\frac{-17}{25}(7,1)= \frac{1}{25}(-31,217)[/tex]

By construction, we have that w+b=u. We need to check that they are orthogonal. To do so, the dot product between w and b must be zero. Recall that if we have vectors a,b  that are orthogonal then for every non-zero escalar r,k the vector ra and kb are also orthogonal. Then, we can check if w and b are orthogonal by checking if the vectors (7,1) and (-31, 217) are orthogonal.

We have that (7,1)\cdot(-31,217) = 7\cdot -31 + 217 \cdot 1 = -217+217 =0. Then this vectors are orthogonal, and thus, w and b are orthogonal.

5 0
3 years ago
Other questions:
  • Please help 100 points! Will mark Brainliest. Thank You!
    14·2 answers
  • Round 372.87 to the nearest TENTH.
    6·2 answers
  • Please help!!!!
    8·1 answer
  • Write a verbal sentence to represent x cubed-9=27
    5·1 answer
  • What is half of 1 2/3 cups
    7·1 answer
  • If matt's average speed was 8 km/hr how far did he travel in 15 minutes
    10·1 answer
  • Pls pls help it's really urgent!!! ​
    10·1 answer
  • Ten deer were introduced into a protected preserve. Their expected population doubles each year. Which function represents the s
    12·1 answer
  • In a certain year, the price of gasoline rose by 20% during January, fell by 20% during February, rose by 25% in March, and fell
    8·1 answer
  • The value of <img src="https://tex.z-dn.net/?f=%5Csin%20%5E%7B-1%7D%5Cleft%28%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5Cright%29-%5Csi
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!