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3 years ago

Which of the h-values satisfy the following inequality?

Mathematics

2 answers:

olasank [31]3 years ago

6 0

Answer:

Step-by-step explanation:

ElenaW [278]3 years ago

6 0

The answer I got for this question was 12

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Approximately three out of every 25 Americans live in California about three out of every 50 Americans live in New York in about

pickupchik [31]

3 out of 25 = 12 out of 100
3 out of 50 = 6 out of 100
2 out of 25 = 8 out of 100
12 + 6 + 8 = 26
100 - 26 = 74%

4 0

3 years ago

sarah had 3 cookies in her jar her friend angela took -18 from her jar how many cookies does she have now?

s344n2d4d5 [400]

She gonna owned like 20

3 0

4 years ago

Read 2 more answers

For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to thi

julsineya [31]

Answer:

1. (-4,6) there is no a solution to the equation through this point

2. (2,−6) there is no a solution to the equation through this point

3. (−5,39) there is a solution to the equation through this point

4. (−1,45) there is a solution to the equation through this point

Step-by-step explanation:

Using the existence and uniqueness theorem:

Let:

$F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }$

Now, let's find the domain of $F(x,y)$ , due to the square root:

$y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6$

So the domain of the function is:

$y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6$

Now, due to the fraction $\frac{\partial F}{\partial y}$ the denominator must be also different from 0, so:

$y^2-36\neq0\\\\y \neq \pm6$

So, the theorem tells us that for each $y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0$ there exists a unique solution defined in an open interval around $x_0$ .

1. (-4,6) there is no a solution to the equation through this point because $y_0=6$

2. (2,−6) there is no a solution to the equation through this point because

$y_0=-6$

3. (−5,39) there is a solution to the equation through this point because

$y_0>6$

4. (−1,45) there is a solution to the equation through this point because

$y_0>6$

6 0

3 years ago

Tony rounded each of the numbers 1, 143 and 1, 149 to the nearest hundred which word correctly compares the rounding numbers

Montano1993 [528]

Full Question:

Tony rounded each of the numbers 1,143 and 1,149 to the nearest hundred. Which choice correctly compares the rounded numbers?

$1,000 = 1,000$

$1,140< 1,150$

$1,100 = 1.100$

$1,140>1,150$

Answer:

$1,100 = 1,100$

Step-by-step explanation:

Given

1,143 and 1,149

Required

Which of the option is correct

We start by approximating both numbers to nearest digit

1,143; when approximated to nearest hundred is 1,100

1,149; when approximated to nearest hundred is also 1,100

Hence;

1,143 ≅ 1,100

1,149 ≅ 1,100

Comparing both results, we have that

$1,100 = 1,100$

From the list of given options, option C is correct;

4 0

4 years ago

Help please..... I cant find the answe

k0ka [10]

Answer:

$6^{30}$

Step-by-step explanation:

Using the rules of exponents

$(a^m)^{n}$ ⇔ $a^{mn}$

$\frac{a^{m} }{a^{n} }$ ⇔ $a^{(m-n)}$

Given

$\frac{(6^-4)^{-9} }{6^6}$

= $\frac{6^{36} }{6^{6} }$

= $6^{(36-6)}$

= $6^{30}$

5 0

3 years ago