All categories

3 years ago

-9 x 2 1/2 What is the value of the expression

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

3 0

Answer:

i think its -22.5

Step-by-step explanation:

i just multiplied -9 x 2 1/2 and got -22.5

You might be interested in

7(2x+5) = 11x-9-x <br><br> Please HELP

mr_godi [17]

Step-by-step explanation:

7(2x+5)=11x-9-x

14x+35=10x-9

4x+35=-9

4x=-44

x=-11

3 0

3 years ago

Need the steps too this

ruslelena [56]

The value of x is 9.

We have : - 25 = 7 - (5x - 13)

We have to find the value of x.

<h3>If f(x) = g(x), than what do you understand by true solution of this equation?</h3>

The true Solution of the equation is the value of x for which LHS = RHS.

For example - if ax + b = cx + d , then

ax - cx = d - b

x = $\frac{d-b}{a-c}$ is the true solution of the equation.

In the question given -

- 25 = 7 - (5x - 13)

- 25 = 7 - 5x + 13

5x = 45

x = 9

Hence, the value of x for which LHS = RHS is 9.

To solve more questions on finding unknown variable, visit the link below -

brainly.com/question/6866597

#SPJ1

3 0

2 years ago

Please help with the question above :)!!! ^^

zhannawk [14.2K]

(-1,-9) because if we substituted 1 into x y would be 9 and if we substituted 1/2 into x y would be 3 but if we substituted -1 into x y would be 1/9 so therefore that value is incorrect

7 0

3 years ago

Clara writes the equation (x – 13)(x + 8) = 196 to solve for the missing side length of a triangle represented by the factor x +

Kitty [74]

<span>(x – 13)(x + 8) = 196
</span>x * x + x * 8 - 13 * x - 13 * 8 = 196
x² + 8x - 13x - 104 = 196
x² - 5x - 104 - 196 = 0
x² - 5x - 300 = 0

The missing side length is 28

3 0

3 years ago

2. Calculate an expression for dy/dx and d2y/dx2 in terms of t if the parametric pair is given as tan(x) = e^at and e^y = 1 + e^

Ber [7]

I assume a is a constant. If tan(x) = exp(at) (where exp(x) means eˣ), then differentiating both sides with respect to t gives

sec²(x) dx/dt = a exp(at)

Recall that

sec²(x) = 1 + tan²(x)

Then we have

(1 + tan²(x)) dx/dt = a exp(at)

(1 + exp(2at)) dx/dt = a exp(at)

dx/dt = a exp(at) / (1 + exp(2at))

If exp(y) = 1 + exp(2at), then differentiating with respect to t yields

exp(y) dy/dt = 2a exp(2at)

(1 + exp(2at)) dy/dt = 2a exp(2at)

dy/dt = 2a exp(2at) / (1 + exp(2at))

By the chain rule,

dy/dx = dy/dt • dt/dx = (dy/dt) / (dx/dt)

Then the first derivative is

dy/dx = (2a exp(2at) / (1 + exp(2at))) / (a exp(at) / (1 + exp(2at))

dy/dx = (2a exp(2at)) / (a exp(at))

dy/dx = 2 exp(at)

Since dy/dx is a function of t, if we differentiate dy/dx with respect to x, we have to use the chain rule again. Suppose we write

dy/dx = f(t)

By the chain rule, the derivative is

d²y/dx² = df/dx

d²y/dx² = df/dt • dt/dx

d²y/dx² = (df/dt) / (dx/dt)

d²y/dx² = 2a exp(at) / (a exp(at) / (1 + exp(2at)))

d²y/dx² = 2 (1 + exp(2at))

4 0

2 years ago