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3 years ago

Without graphing, where will the line y=7x-2 cross the y axis?

Mathematics

1 answer:

Vadim26 [7]3 years ago

5 0

Answer:

The line crosses the y-axis at -2

Step-by-step explanation:

y = mx + b 'm is the slope and b is the y-intersect (the place where the graph comes across the y-axis.

y = mx + b; y = mx + b; b = -2.

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ElenaW [278]

Answer:

acute angle

Step-by-step explanation:

<h2>Finding the angle </h2>

an obtuse angle is an angle which is more than 90° but less than 180°

a right angle is 90°

a straight angle is 180°

an acute angle is less than 90°

is you look at the picture and the definitions you can notice that the angle is less than 90 ° hence its an acute angle

7 0

3 years ago

Read 2 more answers

This a 2 part question. F(x) is the graph shown.

Vlada [557]

Suppose f(x) is an even function. The range of g(x) = -3f(2x) is [-9, 3].

Suppose f(x) is an odd function. What is the range of 0.5f(x-3) is [-1.5, 0.5]

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3 years ago

8x+4+66666+7+5+5+%3+3

anastassius [24]

Answer:

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3 0

3 years ago

Hello I need help Finding the probability

Stolb23 [73]

Each petal of the region $R$ is the intersection of two circles, both of diameter 10. Each petal in turn is twice the area of a circular segment bounded by a chord of length $5\sqrt2$ , which implies the segment is subtended by an angle of $\dfrac\pi2$ . This means the area of the segment is

$\text{area}_{\text{segment}}=\text{area}_{\text{sector}}-\text{area}_{\text{triangle}}$
$\text{area}_{\text{segment}}=\dfrac{25\pi}4-\dfrac{25}2$

This means the area of one petal is $\dfrac{25\pi}2-25$ , and the area of $R$ is four times this, or $50\pi-100$ .

Meanwhile, the area of $G$ is simply the area of the square minus the area of $R$ , or $10^2-(50\pi-100)=200-50\pi$ .

So

$\mathbb P(X=R)=\dfrac{50\pi-100}{100}=\dfrac\pi2-1$
$\mathbb P(X=G)=\dfrac{200-50\pi}{100}=2-\dfrac\pi2$
$\mathbb P((X=R)\land(X=G))=0$ (provided these regions are indeed disjoint; it's hard to tell from the picture)
$\mathbb P((X=R)\lor(X=G))=\mathbb P(X=R)+\mathbb P(X=G)=1$

4 0

4 years ago

how to integrate <img src="https://tex.z-dn.net/?f=e%5E%7B2s%7D%20%2ACos%20%5Cfrac%7Bs%7D%7B4%7D" id="TexFormula1" title="e^{2s}

icang [17]

Answer:

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

Step-by-step explanation:

Step(i):-

Given that $f(s) = e^{2s} cos\frac{s}{4}$

Now integrating

$\int\limits {f(s)} \, ds = \int\limits {e^{2s} cos\frac{s}{4} ds$

By using integration formula

$\int\limits { e^{ax} cos b x dx = \frac{e^{ax} }{a^{2}+b^{2} } ( a cos b x + b sin b x )$

Step(ii):-

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{e^{2s} }{(2)^{2}+(\frac{1}{4}) ^{2} } ( 2 cos (\frac{1}{4} ) s + \frac{1}{4} sin \frac{1}{4} s ))$

= $\frac{e^{2s} }{(4+\frac{1}{16})} ( 2 cos (\frac{1}{4} ) s + \frac{1}{4} sin \frac{1}{4} s ))$

= $\frac{e^{2s} }{(\frac{65}{16} } ( \frac{8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s}{4} ))$

= $16 X\frac{e^{2s} }{65 } ( \frac{8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s}{4} ))$

= $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

Final answer:-

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

6 0

3 years ago