Solve for y. 5(y+5)=-7(2y-1) + 7y

Mathematics

1 answer:

scZoUnD [109]3 years ago

4 0

Answer: y=1.5

Step-by-step explanation: Isolate the variable by dividing each side by factors that don’t contain the variable.

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A and B are two similar shapes the area of shape A is 200cm2 calculate the area of shape B

barxatty [35]

Answer:

Area of shape B = 312.5 cm²

Step-by-step explanation:

Ratio of Area of A : Area of B = 12² : 15²

Given that area of A is 200 cm², let the area of shape B = x

Therefore:

200:x = 12²:15²

200/x = 144/225

Cross multiply

144x = 200*225

144x = 45,000

x = 45,000/144

x = 312.5 cm²

Area of shape B = 312.5 cm²

5 0

2 years ago

Tell whether the lines of the pair of equations are parallel, perpendicular, or neither.

tino4ka555 [31]

First Equation:

$y= -\frac{3}{4} x+2$

Second Equation can be written as:

$y= \frac{3}{4}x+2$

Slope of first equation is -3/4 and slope of second equation is 3/4.
Slope of parallel lines must be equal, and slope of perpendicular lines are the negative reciprocal of each other. None of these conditions can be seen for given two equations.

So, the two lines are neither parallel nor perpendicular.
So correct option is C

7 0

3 years ago

Read 2 more answers

Find the interval(s) of upward concavity on this accumulation function.

maxonik [38]

$\displaystylef(x)=\int_{0}^{x^2}\sec^2(\sqrt{x})dx \\=\int_{0}^{\sqrt{x^2}}\sec^2(u)\cdot2udu \\=2\int_{0}^{\sqrt{x^2}}\sec^2(u)du \\=2\Big[u\tan(u)-\int\tan(u)du\Big]_{0}^{\sqrt{x^2}} \\=2\Big[u\tan(u)+\ln\Big(\mathrm{abs}(\cos(u))\Big)\Big]_0^x \\=2\Big(x\tan(x)+\dfrac{1}{2}\ln\Big(\cos^2(x)\Big)\Big) \\=2x\tan(x)+\ln(\cos^2(x))$