All categories

2 years ago

4.Find the solution mentally.

Mathematics

2 answers:

Trava [24]2 years ago

6 0

3*-4=-12
4*-3=-12
-12*-1=12

garik1379 [7]2 years ago

3 0

Answer:

3×-4=a

a=-12

b×(-3)= -12

-3b= -12

b=-12

-12c=12

c=-1

d×24= -12

24d= -12

d= -1/2 or -0.5

You might be interested in

Pls help it due ASAP. <br> Best get brainiest

natali 33 [55]

Assuming it is a Obtuse Scalene Triangle
Side a = 6
Side b = 15
Side c = 13.0767 = 3√19

5 0

2 years ago

HELPP!!!!!!!!!!!!!!!!

Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step. The equation is

$8-3\sqrt[5]{x^3}=-7$

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time. First thing is to subtract 8 from both sides:

$-3\sqrt[5]{x^3}=-15$

The goal is to isolate the term with the x in it, so that means that the -3 has to go. Divide it away on both sides:

$\sqrt[5]{x^3}=5$

Let's rewrite that radical into exponential form:

$x^{\frac{3}{5}}=5$

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

$(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}$

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely. On the right, let's rewrite that back in radical form to solve it easier:

$x=\sqrt[3]{5^5}$

Let's group that radicad into groups of 3's now to make the simplifying easier:

$x=\sqrt[3]{5^3*5^2}$ because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

$x=5\sqrt[3]{5^2}$ which is the same as:

$x=5\sqrt[3]{25}$

8 0

2 years ago

You are constructing a tangent line from a point outside a given circle to the circle. You have drawn the circle, a point outsid

sveta [45]

Yikes I might go with C.

6 0

2 years ago

Read 2 more answers

How do you do slope intercept

viva [34]

Answer:

on a graph you do rise over run. find a point that falls on an intersecting line and count how many blocks up and over it is to the next point on a line Hope this helped

8 0

2 years ago

Read 2 more answers

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function.

marishachu [46]

Answer:

$\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18$

Step-by-step explanation:

We have been given the function

$f(x)=-2x^2+4x^3+3x^2+18$

From the rational zeros theorem, we have

$\text{Possible rational zeros}=\pm\frac{\text{Factors of constant term}}{\text{Factors of leading coefficient}}$

From the given function,

Leading coefficient = 2

Factors of 2 are 1,2

Constant term = 18

Factors of constant term = 1, 2, 3, 6, 9, 18

Hence, we have

$\text{Possible rational zeros}=\pm\frac{1,2,3,6,9,18}{1,2}\\\\\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18$

6 0

2 years ago