All categories

3 years ago

5x-4[7+(2x-4)], for x=-3 with shown work please

Mathematics

2 answers:

wel3 years ago

5 0

Answer: -3

Concept:

Here, we need to understand the idea of evaluation.

When encountering questions that gave you an expression with variables, then stated: "If x = a, y = b, z = c" (a, b, c are all constants), this means you should substitute the value given for each variable back to the expression.

Solve:

Given information

x= -3

Given expression

5x - 4 [7 + (2x - 4)]

Substitute x value into the expression

=5(-3) - 4 [7 + (2(-3) - 4)]

Simplify by multiplication

=-15 - 4 [7 + (-6 - 4)]

Simplify value in parentheses

=-15 - 4 [7 + (-10)]

Simplify value in bracket

=-15 - 4 [-3]

Simplify by multiplication

=-15 - (-12)

=-15 + 12

= $\boxed{-3}$

Hope this helps!! :)

Please let me know if you have any questions

katovenus [111]3 years ago

5 0

5×(－3)－4{7＋[2×(－3)－4]}
－15－4[7＋(－6－4)]
＝－15－4[7＋(－10)]
＝－15－4(7－10)
＝－15－4×(－3)
＝－15＋12
＝－3

Hope my answer helped u :)

You might be interested in

The volume of a spherical cancerous tumor is given by the following equation. V(r) =(4/3)pi r^3 If the radius of a tumor is esti

Dimas [21]

Answer:

The error that might occur is $\±0.123cm^3$

Step-by-step explanation:

Given

$V = \frac{4}{3}\pi r^3$

$Radius,\ r = 1.4cm$

$Error = 0.005cm$

Required

Determine the error that might occur in the volume of the tumor

Given that there's an error in measurement, this question will be solved using the concept of differentiation;

First, we'll rewrite the given parameters in differentiation notations;

$r = 1.4$ --- Radius

$dV = \±0.005$ --- Change in measurement

$V(r) = \frac{4}{3}\pi r^3$ --- Volume as a function of radius

The relationship between the above parameters is as follows;

$\frac{dV}{dr} = V'(r)$

This can be rewritten as

$\frac{dV}{dr} = (V(r))'$

Substitute $\frac{4}{3}\pi r^3$ for $V(r)$

$\frac{dV}{dr} = (\frac{4}{3}\pi r^3)'$

Multiply both sides by dr

$dr * \frac{dV}{dr} = (\frac{4}{3}\pi r^3)' * dr$

$dV = (\frac{4}{3}\pi r^3)' * dr$

$dV = \frac{4}{3}\pi( r^3)' dr$

Differentiate

$dV = \frac{4}{3}\pi( r^3)' dr$

$dV = \frac{4}{3}\pi* 3r^2\ dr$

$dV = 4\pi* r^2\ dr$

Substitute the values of r, dr and take $\pi$ as 3.142

$dV = 4 * 3.142* 1.4^2 * \±0.005$

$dV = \±0.1231664$

$dV = \±0.123$ (Approximated)

Hence, the error that might occur is ±0.123

6 0

3 years ago

In geometry what is the purpose of sin and cos?

Rasek [7]

Sin cos and tan are basically just functions that relate an angle with a ratio of two sides in a right triangle. Sin is equal to the side opposite the angle that you are conducting the functions on over the hypotenuse which is the longest side in the triangle. Cos is adjacent over hypotenuse.

5 0

3 years ago

Read 2 more answers

Nancy is going to glaze 6-ounce servings of pretzels and put them into gift boxes. If pretzels

Shtirlitz [24]

When in doubt guess C. Always works for me

8 0

3 years ago

Which list of numbers contains an irrational number?

Alinara [238K]

because the 0, if u get calculator u will know what im talking about

5 0