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
MakcuM [25]
3 years ago
7

Help please.. Solve the equation x^2 + 35x = 0

Mathematics
1 answer:
insens350 [35]3 years ago
5 0
X²+35=0
X(x+35)=0

X=0 x+35=0
-35 -35
X=-35

x=0, x=-35
You might be interested in
How do i solve this problem
Vladimir79 [104]
Solve the following system:
{6 t - 5 s = -4 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{-2 r - 4 s - 4 t = -9 | (equation 3)

Swap equation 1 with equation 3:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Subtract 1/2 × (equation 1) from equation 2:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 1 by -1:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 2 by 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 4 s + 10 t = 1 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Swap equation 2 with equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r - 4 s + 10 t = 1 | (equation 3)
Subtract 4/5 × (equation 2) from equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+(26 t)/5 = 21/5 | (equation 3)

Multiply equation 3 by 5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+26 t = 21 | (equation 3)
Divide equation 3 by 26:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s+0 t = (-115)/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 2 by -5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 2) from equation 1:
{2 r + 0 s+4 t = 25/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 3) from equation 1:
{2 r+0 s+0 t = (-17)/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 1 by 2:
{r+0 s+0 t = (-17)/26 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
v0 r+0 s+t = 21/26 | (equation 3)
Collect results:Answer:  {r = -17/26
               {s = 23/13                        {t = 21/26
7 0
3 years ago
Helppp!!!! please!!!
Vitek1552 [10]

Answer:

d. 15 square yard

Step-by-step explanation:

area \: of \: shape \\  =  \frac{1}{2}  \times base \times height \\  \\  =  \frac{1}{2}  \times 10 \times 3 \\  \\  =  \frac{1}{2}  \times 30 \\  \\  = 15 \:  {yd}^{2}

4 0
3 years ago
Read 2 more answers
(2p + 4) + 5 (p-1) - (p+7)
jonny [76]

<em>Your answer will be, </em><em>"6p - 8"</em>

Thanks,

<em>Deku ❤</em>

6 0
3 years ago
What is the answer 4d+12e=
Phantasy [73]

Answer:

d = -3e

Step-by-step explanation:

hope this helps

8 0
3 years ago
What are the slopes of the two lines? <br> green has two yellow dots.
IrinaVladis [17]

Answer:

-7(1/7) = -1

Step-by-step explanation:

4 0
3 years ago
Other questions:
  • Is (-5,-5) a solution of y \geq -2x +4y≥−2x+4 ?
    15·1 answer
  • 2(x+7)+3(x+1) what is the answer
    14·1 answer
  • What is the value of T(-4) when T(x) = -x - 2
    13·1 answer
  • Relationship B has a greater rate than Relationship A. This graph represents Relationship A.
    8·1 answer
  • Differentiating ln inside a bracket
    8·1 answer
  • Jessica has 84.5 of fabric to make curtain
    15·1 answer
  • HELP I RLLY NEED IT
    15·1 answer
  • HELP I NEED HELP ASAP
    7·1 answer
  • The answer is 6 just did it
    5·2 answers
  • Solve 3x≤ -6 graph the sloution
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!