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patriot [66]
3 years ago
10

Ill mark brainlist plss help

Mathematics
1 answer:
IgorC [24]3 years ago
4 0
222 ft jshdhdsvhssdbsh
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Jasmin's dog weighs 55 pounds. Her vet told her that a healthy weight for her dog would be less than or equal to 42 pounds. Jasm
Mnenie [13.5K]

Answer:

x = 2

Step-by-step explanation:

55-42=13

13 brought up to 14 so it can be divided by 7

14 / 7 = 2

8 0
4 years ago
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Find the value of k given that logk - log (k-2) = log5
victus00 [196]

k = \frac{5}{2}

using the ' laws of logarithms '

log x - log y = log ( \frac{x}{y})

logx = logy ↔ x = y

logk - log(k - 2) = log5

log ( \frac{k}{(k - 2)} =log 5

\frac{k}{(k - 2)} = 5

5(k - 2) = k

5k - 10 = k

4k = 10 ⇒ k = \frac{10}{4} = \frac{5}{2}


3 0
4 years ago
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An isosceles right-angled triangle is shown.
Bad White [126]

Answer:

\boxed{x=4}

Step-by-step explanation:

Area of triangle = 0.5 × Base × Height

→ Substitute in the values

200 = 0.5 × 5x × 5x

→ Simplify

200 = 12.5x²

→ Divide both sides by 12.5

16 = x²

→ Square root both sides

4 = x ∴ x = 4

6 0
3 years ago
(a) The plane y + z = 13 intersects the cylinder x2 + y2 = 25 in an ellipse. Find parametric equations for the tangent line to t
klemol [59]

Answer:

Step-by-step explanation:

We have a curve (an ellipse) written as the system of equations

\begin{cases} y+z &= 13\\ x^2+y^2 &= 25\end{cases}.

And we want to calculate the tangent at the point (3,4,9).

The idea in this problem is to consider two variables as functions of the third. Usually we consider y and z as functions of x. Recall that a curve in the space can be written in parametric form in terms of only one variable. In this case we are considering the ‘‘natural’’ parametrization (x, y(x), z(x)).

Recall that the parametric equation of a line has the form

r(t)=\begin{cases} x(t) &= x_0 + v_1t \\ y(t) &= y_0 +v_2t\\ z(t) &= z_0 +v_3t \end{cases},

where (x_0,y_0,z_0) is a point on the line (in this particular case is (3,4,9)) and (v_1,v_2,v_3) is the direction vector of the line. In this case, the direction vector of the line is the tangent vector of the ellipse at the point (3,4,9).

Now, if we have the parametric equation of a curve (x, y(x), z(x)) its tangent line will have direction vector (1, y'(x), z'(x)). So, as we need to calculate the equation of the tangent line at the point (3,4,9) = (3, y(3), z(3)), we must obtain the tangent vector (1, y'(3), z'(3)). This part can be done taking implicit derivatives in the systems that defines the ellipse.

So, let us write the system as

\begin{cases} y(x)+z(x) &= 13\\ x^2+y^2(x) &= 25\end{cases}.

Then, taking implicit derivatives:

\begin{cases} y'(x)+z'(x) &= 0 \\ 2x+2y(x)y'(x) &= 0\end{cases}.

Now we substitute the values x=3 and y(3)=4, and we get the system of linear equations

\begin{cases} y'(3)+z'(3) &= 0 \\ 2\cdot 3+2\cdot 4y'(x) &= 0\end{cases},

where the unknowns are y'(3) and z'(3).

The system is

\begin{cases} y'(3)+z'(3) &= 0 \\ 6+8y'(x) &= 0\end{cases},

and its solutions are

y'(3) = -\frac{3}{4} and z'(3) = \frac{3}{4}.

Then, the direction vector of the tangent is

(1, -\frac{3}{4}, -\frac{3}{4}).

Finally, the tangent line has parametric equation

r(t)=\begin{cases} x(t) &= 3 + t \\ y(t) &= 4 -\frac{3}{4}t\\ z(t) &= 9 +\frac{3}{4}t \end{cases}

where t\in\mathbb{R}.

7 0
4 years ago
I'm having trouble understanding this word problem and how they got 6x=44+2x from it.
jenyasd209 [6]
Ok well since it says the product of a number and 6 is 44 more than twice the number.  and from this they got 6x=44+2x because it SAYS THE PRODUCT OF A NUMBER U DON'T KKNOW WHAT THE NUMBER IS SO U USE A VARIABLE WHICH IS 6X THEY GOT = 44 FROM THE PRODUCT AND 2X FROM MORE THAN TWICE THE NUMBER AND TWICE IS 2 AND U DON'T KNOW WHAT THE NUMBER IS SO ONCE AGAIN USE A VARIABLE  
                   
 HOPE THIS HELPED
5 0
3 years ago
Read 2 more answers
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