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natita [175]
3 years ago
14

Help needed ASAP will give brainliest! Pls help me

Mathematics
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

1740 N

Step-by-step explanation:

To find the force, you need to multiply the mass by the acceleration.

(60.0 kg) × (29.0 m/s²) = 1740 N

The force will be 1740 N.

You might be interested in
Mark -5/7, 2/7, 6/7, -1 on a number line.<br>if u will answer i will mark u as brainlist.​
DerKrebs [107]

Answer:

So first converthem into a deciaml if the number line is ordered by decimals or make them from least to greatest, left to right.

Decimal:

If asking for s decimal, the -1 is definitely the least so -1. For -5/7, it’s just -0.71 (rounded), for 2/7, its 0.28 and for 6/7 its 0.86.

6 0
3 years ago
Identify the zero(s) of the function graphed below
k0ka [10]

Answer:

Step-by-step explanation:

The zeros are the values of x for which y=0.

The zero of this graph is x=1.

8 0
3 years ago
Find the roots of:
Elena-2011 [213]

\large\displaystyle\text{$\begin{gathered}\sf \pmb{1) \  2x^3-7x^2+8x-3=0 } \end{gathered}$}

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

  | 2  -7    8  -3

<u>1 |      2   -5   3</u>

  | 2   -5    3  0

<u> 1 |      2     -3   </u>

    2   -3     0

So the factorization is (x-1)² (2x-3)=0. So:

                     \bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2}  }

\large\displaystyle\text{$\begin{gathered}\sf \pmb{2) \  x^3-x^2-4=0 } \end{gathered}$}

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

      |  1  -1  0  -4

  <u>2  |     2  2     </u>

         1  2  2  0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

                   \bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }

\large\displaystyle\text{$\begin{gathered}\sf \pmb{3) \  6x^3+7x^2-9x+2=0 } \end{gathered}$}

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

   | 6    7       9      2

<u>-2 |      -12    10     -2</u>

     6    -5     1       0

So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

         \bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}

                     \large\displaystyle\text{$\begin{gathered}\sf  \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered}$}

6 0
2 years ago
Find all real zeros of f(x) = 2x^2 +5x-18 algebraically.
den301095 [7]
2x2-5x-18=0

Two solutions were found :

x = -2
x = 9/2 = 4.500
Step by step solution :

Step 1 :

Equation at the end of step 1 :

(2x2 - 5x) - 18 = 0
Step 2 :

Trying to factor by splitting the middle term

2.1 Factoring 2x2-5x-18

The first term is, 2x2 its coefficient is 2 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is -18

Step-1 : Multiply the coefficient of the first term by the constant 2 • -18 = -36

Step-2 : Find two factors of -36 whose sum equals the coefficient of the middle term, which is -5 .

-36 + 1 = -35
-18 + 2 = -16
-12 + 3 = -9
-9 + 4 = -5 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -9 and 4
2x2 - 9x + 4x - 18

Step-4 : Add up the first 2 terms, pulling out like factors :
x • (2x-9)
Add up the last 2 terms, pulling out common factors :
2 • (2x-9)
Step-5 : Add up the four terms of step 4 :
(x+2) • (2x-9)
Which is the desired factorization

Equation at the end of step 2 :

(2x - 9) • (x + 2) = 0
Step 3 :

Theory - Roots of a product :

3.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.
4 0
3 years ago
Read 2 more answers
At what point is the following function continuous?
liubo4ka [24]

Check the picture below.

something worth noticing           \bf \cfrac{x^2-x-6}{x-3}\implies \cfrac{(\underline{x-3})(x+2)}{\underline{x-3}}\implies x+2

so, we're really graphing x+2, with a hole at x = 3, however, when x = 3, we know that f(x) = 5, but but but, when x = 3, x+2 = 5, so we end up with a continuous line all the way, x ∈ ℝ, because the "hole" from the first subfunction, gets closed off by the second subfunction in the piece-wise.

4 0
3 years ago
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