All categories

2 years ago

Which is greater, -11 or -9 ?

Mathematics

2 answers:

Vitek1552 [10]2 years ago

7 0

Answer:

-9

Step-by-step explanation:for sure

ElenaW [278]2 years ago

7 0

Answer:

-9

Step-by-step explanation:

Look at a number line:

<----+------+------+------+------+------+------+------+------+------+------+------+------+------+-->

-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1

On a number line, a number to the right is always greater than a number to its left. Just like 1 is greater than 0, -9 is greater than -11.

-9 > -11

Answer: -9

You might be interested in

HELPPPPPPPPPPPPPPPPPPPPPPP PLSSSSSS

Natasha2012 [34]

Answer:

values of k:

any real number works

Step-by-step explanation:

k^2 is always >=0 so

k^2+96 is always >0

7 0

2 years ago

Find the sum using estimation to the nearest dollar: $72.90+$15.45+$120.99

o-na [289]

Answer: 209

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

the cable of a hoist has a cross-section of 80 mm2. the hoist is used to lift a crate weighing 500 kg. what is the stress in the

Daniel [21]

Answer:

stress on the cable is 61.312mpa

If steel is used, extension will be 0.91968mm

If polypropylene is used, extension will be 0.1532mm

Step-by-step explanation: detailed explanation and calculation is shown in the image below

3 0

2 years ago

In abc above,what is the length of ad

Nezavi [6.7K]

Answer:

Step-by-step explanation:

First calculate BD using sine ratio in Δ BCD and the exact value

sin60° = $\frac{\sqrt{3} }{2}$ , thus

sin60° = $\frac{opposite}{hypotenuse}$ = $\frac{BD}{BC}$ = $\frac{BD}{12}$ = $\frac{\sqrt{3} }{2}$ ( cross- multiply )

2BD = 12 $\sqrt{3}$ ( divide both sides by 2 )

BD = 6 $\sqrt{3}$

-----------------------------------------------------------

Calculate AD using the tangent ratio in Δ ABD and the exact value

tan30° = $\frac{1}{\sqrt{3} }$ , thus

tan30° = $\frac{opposite}{adjacent}$ = $\frac{AD}{BD}$ = $\frac{AD}{6\sqrt{3} }$ = $\frac{1}{\sqrt{3} }$ ( cross- multiply )

$\sqrt{3}$ AD = 6 $\sqrt{3}$ ( divide both sides by $\sqrt{3}$ )

AD = 6 → B

4 0

2 years ago