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alekssr [168]
3 years ago
8

Instead of subtracting 10, you can add________. (Complete the sentence)

Mathematics
1 answer:
Fiesta28 [93]3 years ago
4 0
Instead of subtracting 10, you can add negative 10
You might be interested in
All zeros of polynomial function 3x^4 +14x^2 -5
Marizza181 [45]

Answer:

Correct answer:  x₁ = 1 / √3 = √3 / 3  or  x₂ = - 1 / √3 = - √3 / 3

Step-by-step explanation:

Given:

3 x⁴ + 14 x² - 5 = 0   biquadratic equation

this equation is solved by a shift  x² = t and get:

3 t² + 14 t - 5 = 0

t₁₂ = (-14 ± √14² - 4 · 3 · 5) / 2 · 3 = (-14 ± √196 + 60) / 6

t₁₂ = (-14 ± √256) / 6 = (-14 ± 16) / 6

t₁ = -5   or  t₂ = 1 / 3

the solution t₁ = -5 is not accepted because it cannot be x² = -5

we accepted  t₂ = 1 / 3

x² =  1 / 3  ⇒  

x₁ = 1 / √3 = √3 / 3  or  x₂ = - 1 / √3 = - √3 / 3

God is with you!!!

5 0
3 years ago
#5 and #6 plz Will give Brainlyest!
Evgesh-ka [11]

Answer:

5. 3.93833333333 rounded: 3.94

6.Start with 1.8, then add 1.05 repeatedly or Start with 1.8, then add 1.05 repeatedly

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Evaluate-2+7x^2-2x^3-8x+7x^4 at x=5
Murljashka [212]

Answer:

4262

Step-by-step explanation:

2 + 7x^2- 2x^3-8x +7x^4     at x = 5

= 2 + 7*(5)^2 - 2*(5)^3-8*5 +7*(5)^4

=2+175-250-40+4375

=4552-290

=4262

4 0
3 years ago
A square pyramid is shown. What is the surface area?
Ber [7]

Answer:

The total surface area of a regular pyramidis the sum of the areas of its lateralfaces and its base. The general formula for thelateral surface area of a regular pyramidis where p represents the perimeter of thebase and SI the height of inclination.

5 0
3 years ago
An equation that represents the path of a diver jumping off a diving board is y = –7x2 + 5x + 16. Graph this function from the p
ANTONII [103]

Answer:

See explanation

Step-by-step explanation:

An equation that represents the path of a diver jumping off a diving board is

y = -7x^2 + 5x + 16.

The diver jumped to the water when x = 0.

Then

y=-7\cdot 0^2+5\cdot 0+16=16

The jumper reached the water when y = 0, then

-7x^2+5x+16=0\\ \\D=5^2-4\cdot (-7)\cdot 16=25+448=473\\ \\x_{1,2}=\dfrac{-5\pm\sqrt{473}}{-14}

So, point where the  jumper reached the water is

\left(\dfrac{5+\sqrt{473}}{14},0\right)\approx (1.93,0)

7 0
3 years ago
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