All categories

3 years ago

Shyla had a balance of −$255.25 in her account, and Susan had a balance of −$178.75 in her account.

Mathematics

2 answers:

IgorLugansk [536]3 years ago

7 0

Answer:

Shyla had a balance of −$255.25 in her account, and Susan had a balance of −$178.75 in her account.

Which statement explains why Shyla owes the bank more than Susan?

−255.25 < −178.75

−255.25 > −178.75

|−255.25| > |−178.75|

|−255.25| < |−178.75| C is the answer

Step-by-step explanation:

I just took a test and C was right

densk [106]3 years ago

4 0

Your answer is A. When doing negatives, the smaller number is actually greater.
Hope this Helps!
A thanks and/or brainliest would be appreciated :D

You might be interested in

Need help ASAP !!!!!!

Slav-nsk [51]

Answer:

Therefore, Point M( 1 , -2 ) is the Mid point of segment ST.

Step-by-step explanation:

Given:

Let,

point S( x₁ , y₁) ≡ ( -1 , 1)

point T( x₂ , y₂) ≡ (3 , -5)

Point M( x , y ) is the Mid point of segment ST.

To Find:

Point M( x , y )= ?

Solution:

As Point M( x , y ) is the Mid point of segment ST.

So we have Mid Point Formula as

$Mid\ pointM(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

On substituting the given values in above equation we get

$M(x,y)=(\frac {-1+3}{2},\frac{-5+1}{2})\\\\M(x,y)=(\frac{2}{2},\frac{-4}{2}) \\\\M(x,y)=(1,-2)\ \textrm{which the required midpoint}$

Therefore, Point M( 1 , -2 ) is the Mid point of segment ST.

4 0

3 years ago

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago

15. Gwen had a painting that had a length of 5 feet and a width of 3.5 feet.

victus00 [196]

Answer:

D. 15 by 10.5

Step-by-step explanation:

If the original painting has dimensions of 5 by 3.5, and has a scale factor of 1, then if we are trying to find the dimensions of the same thing but with a scale factor of 3, then we need to multiply the dimensions by 3.

5*3 = 15

3.5*3 = 10.5

The new dimensions are 15 by 10.5.

Hope this helps. Have a nice day.

5 0

3 years ago

Please help ASAP

Ivan

Answer:

(5, 12, 13)

Step-by-step explanation:

5^2+12^2=25+144=169=13^2

5^2+12^2=13^2

5 0

2 years ago

Calculate the vapor pressure (in torr) of a solution containing 0.2 moles of solute dissolved in 10 moles of water at 25°C. Vapo

Gnesinka [82]

Answer:

the vapor pressure of a solution at 25°C that contains 76.6 g of glucose (180.2 g/mole) in 250.0 mL of water. The vapor pressure of pure water at 25°C

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers