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3 years ago

Determine the area of the triangular ABC with AC =5cm and height DB=30mm

Mathematics

1 answer:

seropon [69]3 years ago

4 0

Answer:

Step-by-step explanation:

AC² = AD² + DB ²

25 = AD² + 9

AD²= 25 - 9

AD² = 16

AD = 4

Area = 4*3 /2

= 6cm

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Marat540 [252]

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$-8x^3-64x^2-64=64-64$

$-8x^3-64x^2-64=0$

$-8x^2+0.97035\dots x-7.88051\dots \approx \:0$

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2) $4n+\frac{3}{2}=2n-\frac{5}{1}$

$4n+\frac{3}{2}=2n-5$

$4n+\frac{3}{2}-\frac{3}{2}=2n-5-\frac{3}{2}$

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3 years ago

Calculate the area of each shape below. Figures are not drawn to scale.

STatiana [176]

Answer:

The area of the triangle is 168 meters

Step-by-step explanation:

Formula to find the area of a triangle:

Area = 1/2 Base x Height

So, first you need to identify the base and the height. The base is 16m + 8m, and the height is 14m. All you have to do is plug in the numbers:

Area = 1/2 (16 + 8) x 14

Then, you solve:

Area = 1/2 (16 + 8) x 14

Area = 1/2 24 x 14

Area = 1/2 336

Area = 168 meters

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3 years ago

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4 years ago

Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/

SVEN [57.7K]

Given $p,q>0$ and $\log_9p=\log_{12}q=\log_{16}(p+q)=x$ (say).

Then,

$p=9^x\\ q=12^x\\ p+q=16^x$

From the above 3 equations,

$\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\$

From the equations, we get

$\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}$

Since $p,q>0$ , the negative value is rejected.

$\frac{q}{p}=\frac{1 + \sqrt{5}}{2}$

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4 years ago

Questions 33-36 I don’t know how to find the answer and I need help on how I can find the answer

bulgar [2K]

The answer is 90, 25, 155, 115 respectively

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3 years ago

Determine the area of the triangular ABC with AC =5cm and height DB=30mm​

Determine the area of the triangular ABC with AC =5cm and height DB=30mm