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3 years ago

What is the value of x if AP and BP are tangent segments AP = 10x - 15 and BP = 7x + 3

1 answer:

8 0

If AP and BP are tangent segments connected outside of the circle at point "P" then AP = BP so,
10x - 15 = 7x + 3
3x = 18
x = 6

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Can you help me out with this one?

shepuryov [24]

Answer: 98

Step-by-step explanation:

98 / (-7) = -14

7 0

3 years ago

Read 2 more answers

I WILL AWARD BRAINLIEST!!

AnnZ [28]

Answer:

Can not be determined (CNBD)

Step-by-step explanation:

In the given triangle ACI,

M is the mid point of CI, So we have

CM= MI

also AM=AM ( reflexive property )

So far we have two pairs of corresponding sides are congruent in ΔCAM and ΔIAM

To prove that ΔCAM≅ΔIAM using SSS (side side side ) congruence theorem

we should have AC =AI , but it is not given,so we can not say that AC=AI

To prove that ΔCAM≅ΔIAM using SAS (side angle side ) congruence theorem

we should have ∠AMC=∠AMI, but it is not given,so we can not say that ∠AMC=∠AMI

We can not determine that ΔCAM is congruent to which triangle

Hence the answer is Can not be determined (CNBD)

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3 years ago

Please help. Urgent.

Jet001 [13]

3. ΔPQR ≅ ΔSRT

3. ASA (Angle - Side - Angle) - we have two triangles where we know two angles and the included side are equal

If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

4. PR ≅ SR

4. ΔPQR ≅ ΔSRT - the corresponding sides are congruent.

7 0

3 years ago

Please please please help and solve this with steps, much help needed thank you :) 20 points for this!

leonid [27]

Answer:

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

$\frac{dy}{dx}y=x^3+2x-5x+3$

$\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}$

$\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)$

1. $\mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y'\:y=x^3+2x-5x+3$

2. $\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}$

$yy'\:=x^3-3x+3$

$N\left(y\right)\cdot y'\:=M\left(x\right)$

$N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3$

3. $\mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1$

4. $\mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}$

5. $\mathrm{Simplify}$

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

7 0

3 years ago

Read 2 more answers

What is the equation of the line that is parallel to the given line and passes theought the point (12, -2)

Nesterboy [21]

Answer:

Option D (y = 5/6x -12).

Step-by-step explanation:

Hey there!

The equation of the line which passes through the point (12,-2) is (y+2) = m2(x-12)………(i) [Using one point formula].

According to the question, the first line passes through point (12,6) and (0,-4).

So,

$slope(m1) = \frac{y2 - y1}{x2 - x1}$

$or \: m1 = \frac{ - 4 - 6}{0 - 12}$

$or \: m1 = \frac{5}{6}$

Therefore, the slope of the line is 5/6.

Now as per the condition of parallel lines, m1 =m2 = 5/6.

So, keeping the value of m2 in equation (i), we get;

(y+2) = 5/6(x-12)

$y + 2 = \frac{5}{6} x - 10$

or, y = 5/6x - 12.

Therefore, the required equation is y = 5/6 X - 12.

Hope it helps!

3 0

3 years ago