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3 years ago

Can someone please help me with this question

Mathematics

1 answer:

aniked [119]3 years ago

6 0

2/9 divided by 1/2. invert the second fraction and multiply

2/9 * 2/1 = 4/9

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51. (08.02 MC) Box A has a volume of 48 cubic meters. Box B is similar to box A. To create Box B, Box A's dimensions were double

Vanyuwa [196]

Answer:

384 m³

Step-by-step explanation:

Let the dimension of box a be l,b, and h.

ATQ,

lbh = 48 .....(1)

If we double the dimension of box A, we get box B whose new dimensions will be 2l,2b and 2h.

Let V be the volume of box B.

V = (2l)(2b)(2h)

V = 8(lbh)

= 8(48) [from equation (1)]

= 384 m³

Hence, the volume of the box B is 384 m³.

7 0

2 years ago

Their win percentage was 40, then how many matches did they play in all!

Neko [114]

Answer:

14400

Step-by-step explanation:

X - X × 75%=3600

X - X × $\frac{75}{100}$ = 3600

X - $\frac{3X}{4}$ = 3600

$\frac{4X-3X}{4}$ = 3600

$\frac{X}{4}$ = 3600

X = 3600 × 4

X = 14400

3 0

2 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

How to answer this question??? <br>thanks

vesna_86 [32]

$\dfrac{2m}{5}+3=7(4m-3)\ \ \ \ |use\ distributive\ property:a(b-c)=ab-ac\\\\\dfrac{2m}{5}+3=28m-21\ \ \ \ \ |multiply\ both\ sides\ by\ 5\\\\2m+15=140m-105\ \ \ \ |subtract\ 140m\ from\ both\ sides\\\\-138m+15=-105\ \ \ \ |subtract\ 15\ from\ both\ sides\\\\-138m=-120\ \ \ \ |divide\ both\ sides\ by\ (-138)\\\\m=\dfrac{120:6}{138:6}\\\\\boxed{m=\frac{20}{23}}$

7 0

3 years ago

If two congruent angles form a linear pair, then ___.

Assoli18 [71]

They are acute angles.

6 0

3 years ago

Read 2 more answers