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Sphinxa [80]
3 years ago
9

Does any one know the answer?

Mathematics
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

D is the answer (31.05)

Step-by-step explanation:

2.07×15 is 31.05

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<img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B12x-7%3D0%20%5C%5C" id="TexFormula1" title="2x^{2} +12x-7=0 \\" alt="2x^{2}
m_a_m_a [10]

Answer:

-2. 46

- 9.54

Step-by-step explanation:

  • 2x²+12x-7=0
  • 2x²+12x+18= 25
  • 2*(x²+6x+9)=25
  • 2*(x+6)²= 5²
  • x+6= 5√2/2 ⇒ x= 5√2/2 -6 ≈ -2. 46
  • x+6= -5√2/2 ⇒ x= -5√2/2 -6 ≈ - 9.54
4 0
3 years ago
Find the value of y in the following expression if x = 5 and type the result in the empty box.
ValentinkaMS [17]

Answer:

3.4 is the answer i think

Step-by-step explanation:

3 0
3 years ago
I am lost on what to do
Neko [114]
\bf sin({{ \alpha}})sin({{ \beta}})=\cfrac{1}{2}[cos({{ \alpha}}-{{ \beta}})\quad -\quad cos({{ \alpha}}+{{ \beta}})]&#10;\\\\\\&#10;cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}\\\\&#10;-----------------------------\\\\&#10;\lim\limits_{x\to 0}\ \cfrac{sin(11x)}{cot(5x)}\\\\&#10;-----------------------------\\\\&#10;\cfrac{sin(11x)}{\frac{cos(5x)}{sin(5x)}}\implies \cfrac{sin(11x)}{1}\cdot \cfrac{sin(5x)}{cos(5x)}\implies \cfrac{sin(11x)sin(5x)}{cos(5x)}

\bf \cfrac{\frac{cos(11x-5x)-cos(11x+5x)}{2}}{cos(5x)}\implies \cfrac{\frac{cos(6x)-cos(16x)}{2}}{cos(5x)}&#10;\\\\\\&#10;\cfrac{cos(6x)-cos(16x)}{2}\cdot \cfrac{1}{cos(5x)}\implies \cfrac{cos(6x)-cos(16x)}{2cos(5x)}&#10;\\\\\\&#10;\lim\limits_{x\to 0}\ \cfrac{cos(6x)-cos(16x)}{2cos(5x)}\implies \cfrac{1-1}{2\cdot 1}\implies \cfrac{0}{2}\implies 0
4 0
4 years ago
????????????????? I NEED HELP WITH THIS WILL GIVE BRAINLIEST
GuDViN [60]

Answer:

#3 is 40

#4 is 90

Step-by-step explanation:

3 0
3 years ago
What is the roots of the quadratic equation? 6x^2+5x-4=0
Rama09 [41]

x = \frac{1}{2} or x = - \frac{4}{3}

consider the factors of the product 6 × - 4 = - 24 which sum to the coefficient of the x- term ( + 5)

the factors are + 8 and - 3 ( split the middle term using these factors

6x² - 3x + 8x - 4 = 0 ( factor by grouping )

3x(2x - 1) + 4(2x - 1 ) ( take out common factor of (2x - 1) )

= (2x - 1)(3x + 4) = 0

equate each factor to zero and solve for x

2x - 1 = 0 ⇒ x = \frac{1}{2}

3x + 4 = 0 ⇒ x = - \frac{4}{3}


4 0
4 years ago
Read 2 more answers
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