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2 years ago

If g(x) = 2 − 1, what is g(−2.3)?

Mathematics

2 answers:

jeyben [28]2 years ago

8 0

Answer:g(x) = - 5.6

Step-by-step explanation: first I presume the expression is

g(x) = 2x− 1

Given that g(-2.3)

Substituting this value for x in the expression we have

g(x) = 2(-2.3) − 1

g(x) = - 4.6-1

g(x) = - 5.6

Nadusha1986 [10]2 years ago

6 0

So first you'll substitute x for -2.3 then you'll subtract 2 by 1 which equals 1 then you'll divide -2.3 on both sides which will equal g = -1.3

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Write an inequality for the graph

natima [27]

The open circle means the inequality will be greater than or equal to (≥) or less than or equal to (≤).
A closed circle means the inequality will be greater than (>) or less than (<)

An arrow pointing right to increasingly positive values means the inequality is getting greater (> or ≥)
A narrowing pointing left to increasingly negative values means the inequality is getting lesser (< or ≤)

So for this graph with an open circle and rightward pointing arrow, “x” will be some number on the number line greater than the first point of -38:
x > -38

6 0

2 years ago

Can you help me with these questions

dlinn [17]

Answer:

see below (I hope this helps!)

Step-by-step explanation:

41. √20 = √4 * 5 = √4 * √5 = 2√5

43. The sum of interior angles in a triangle is 180°, therefore, x = 180 - (75 + 64) = 41°.

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47. Complementary angles have a sum of 90°, therefore, x = 90 - 48 = 42°.

7 0

3 years ago

Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

LUCKY_DIMON [66]

Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

8 0

3 years ago

Helppppp me I am stuck help me plzz

soldi70 [24.7K]

Answer:

espero sirva qwq

Step-by-step explanation:

4= 0.56666666666

5=1.66666666667

6=1.375

7=1.425

8=0.92857142857

9=1.666666667

10=1.44444444444

11=0.825

6 0

2 years ago

For what value of k, the zeroes of x2 + kx + 12 will differ by 1?

asambeis [7]

Let a and b be the zeroes of x² + kx + 12 such that |a - b| = 1.

By the factor theorem, we can write the quadratic in terms of its zeroes as

x² + kx + 12 = (x - a) (x - b)

Expand the right side and equate the coefficients:

x² + kx + 12 = x² - (a + b) x + ab

Then

a + b = -k

ab = 12

The condition that |a - b| = 1 has two cases, so without loss of generality assume a > b, so that |a - b| = a - b.

Then if a - b = 1, we get b = a - 1. Substitute this into the equations above and solve for k :

a + (a - 1) = -k → 2a = 1 - k → a = (1 - k)/2

a (a - 1) = 12 → (1 - k)/2 • ((1 - k)/2 - 1) = 12

→ (1 - k)²/4 - (1 - k)/2 = 12

→ (1 - k)² - 2 (1 - k) = 48

→ (1 - 2k + k²) - 2 (1 - k) = 48

→ k² - 1 = 48

→ k² = 49

→ k = ± √(49) = ±7

8 0

2 years ago